The first step to solving this problem is to set the equations equal to each other.

$\displaystyle 51 x^{2} - 46 = 26 x + 27$

Now subtract the right hand side, to make it zero, so we can use the quadratic equation.

$\displaystyle 51 x^{2} - 46 - 26 x - 27 = 0$

$\displaystyle 51 x^{2} - 26 x - 73 = 0$

Recall the quadratic equation.

$\displaystyle x =\frac{ -b \pm\sqrt{ b^{2} - 4 a c }}{ 2 a }$

Where $\displaystyle \uptext{a}$, $\displaystyle \uptext{b}$, and $\displaystyle \uptext{c}$ correspond to coefficients in the following quadratic equation.

$\displaystyle a x^{2} + b x + c = 0$

In this case, $\displaystyle a = 51$ , $\displaystyle b = -26$ , $\displaystyle c = -73$ .

Now plug these values in.

$\displaystyle x =\frac{ 26 \pm\sqrt{\left( -26 \right)^2- 4 \cdot 51 \cdot -73 }}{ 2 \cdot 51 }$

$\displaystyle x =\frac{ 26 \pm 4 \sqrt{973} }{ 102 }$

Split this up into $\displaystyle 2$ equations.

$\displaystyle x =\frac{ 26 + 4 \sqrt{973} }{ 102 }$

$\displaystyle x = 1.48$

$\displaystyle x = - \frac{2 \sqrt{973}}{51} + \frac{13}{51}$

$\displaystyle x = -0.97$

Since we want points, we need to plug these values into one of the original equations.

$\displaystyle p_{1} = 51 \cdot 1.48 + 27$

$\displaystyle p_{1} = 75.48 + 27$

$\displaystyle p_{1} = 102.48$

So the first intersection point is at $\displaystyle \left( 1.48 , 102.48 \right)$

Now we need to find the last point of intersection.

$\displaystyle p_{2} = 51 \cdot -0.97 + 27$

$\displaystyle p_{2} = -49.47 + 27$

$\displaystyle p_{2} = -22.47$

So the second intersection point is at $\displaystyle \left( -0.97 , -22.47 \right)$

The first step to solving this problem is to set the equations equal to each other.

$\displaystyle 24 x^{2} - 70 = 48 x + 93$

Now subtract the right hand side, to make it zero, so we can use the quadratic equation.

$\displaystyle 24 x^{2} - 70 - 48 x - 93 = 0$

$\displaystyle 24 x^{2} - 48 x - 163 = 0$

Recall the quadratic equation.

$\displaystyle x =\frac{ -b \pm\sqrt{ b^{2} - 4 a c }}{ 2 a }$

Where $\displaystyle \uptext{a}$, $\displaystyle \uptext{b}$, and $\displaystyle \uptext{c}$ correspond to coefficients in the following quadratic equation.

$\displaystyle a x^{2} + b x + c = 0$

In this case, $\displaystyle a = 24$ , $\displaystyle b = -48$, $\displaystyle c = -163$ .

Now plug these values in.

$\displaystyle x =\frac{ 48 \pm\sqrt{\left( -48 \right)^2- 4 \cdot 24 \cdot -163 }}{ 2 \cdot 24 }$

$\displaystyle x =\frac{ 48 \pm 4 \sqrt{1122} }{ 48 }$

Split this up into $\displaystyle 2$ equations.

$\displaystyle x =\frac{ 48 + 4 \sqrt{1122} }{ 48 }$

$\displaystyle x = 3.79$

$\displaystyle x = - \frac{\sqrt{1122}}{12} + 1$

$\displaystyle x = -1.79$

Since we want points, we need to plug these values into one of the original equations.

$\displaystyle p_{1} = 24 \cdot 3.79 + 93$

$\displaystyle p_{1} = 90.96 + 93$

$\displaystyle p_{1} = 183.96$

So the first intersection point is at $\displaystyle \left( 3.79 , 183.96 \right)$

Now we need to find the last point of intersection.

$\displaystyle p_{2} = 24 \cdot -1.79 + 93$

$\displaystyle p_{2} = -42.96 + 93$

$\displaystyle p_{2} = 50.04$

So the second intersection point is at $\displaystyle \left( -1.79 , 50.04 \right)$

The first step to solving this problem is to set the equations equal to each other.

$\displaystyle 6 x^{2} - 57 = 43 x + 43$

Now subtract the right hand side, to make it zero, so we can use the quadratic equation.

$\displaystyle 6 x^{2} - 57 - 43 x - 43 = 0$

$\displaystyle 6 x^{2} - 43 x - 100 = 0$

Recall the quadratic equation.

$\displaystyle x =\frac{ -b \pm\sqrt{ b^{2} - 4 a c }}{ 2 a }$

Where $\displaystyle \uptext{a}$, $\displaystyle \uptext{b}$, and $\displaystyle \uptext{c}$ correspond to coefficients in the following quadratic equation.

$\displaystyle a x^{2} + b x + c = 0$

In this case, $\displaystyle a = 6$ , $\displaystyle b = -43$ , $\displaystyle c = -100$ .

Now plug these values in.

$\displaystyle x =\frac{ 43 \pm\sqrt{\left( -43 \right)^2- 4 \cdot 6 \cdot -100 }}{ 2 \cdot 6 }$

$\displaystyle x =\frac{ 43 \pm \sqrt{4249} }{ 12 }$

Split this up into $\displaystyle 2$ equations.

$\displaystyle x =\frac{ 43 + \sqrt{4249} }{ 12 }$

$\displaystyle x = 9.02$

$\displaystyle x = - \frac{\sqrt{4249}}{12} + \frac{43}{12}$

$\displaystyle x = -1.85$

Since we want points, we need to plug these values into one of the original equations.

$\displaystyle p_{1} = 6 \cdot 9.02 + 43$

$\displaystyle p_{1} = 54.12 + 43$

$\displaystyle p_{1} = 97.12$

So the first intersection point is at $\displaystyle \left( 9.02 , 97.12 \right)$

Now we need to find the last point of intersection.

$\displaystyle p_{2} = 6 \cdot -1.85 + 43$

$\displaystyle p_{2} = -11.1 + 43$

$\displaystyle p_{2} = 31.9$

So the second intersection point is at $\displaystyle \left( -1.85 , 31.9 \right)$

The first step to solving this problem is to set the equations equal to each other.

$\displaystyle 9 x^{2} - 89 = 7 x + 24$

Now subtract the right hand side, to make it zero, so we can use the quadratic equation.

$\displaystyle 9 x^{2} - 89 - 7 x - 24 = 0$

$\displaystyle 9 x^{2} - 7 x - 113 = 0$

Recall the quadratic equation.

$\displaystyle x =\frac{ -b \pm\sqrt{ b^{2} - 4 a c }}{ 2 a }$

Where $\displaystyle \uptext{a}$, $\displaystyle \uptext{b}$, and $\displaystyle \uptext{c}$ correspond to coefficients in the following quadratic equation.

$\displaystyle a x^{2} + b x + c = 0$

In this case, $\displaystyle a = 9$, $\displaystyle b = -7$, $\displaystyle c = -113$.

Now plug these values in.

$\displaystyle x =\frac{ 7 \pm\sqrt{\left( -7 \right)^2- 4 \cdot 9 \cdot -113 }}{ 2 \cdot 9 }$

$\displaystyle x =\frac{ 7 \pm \sqrt{4117} }{ 18 }$

Split this up into $\displaystyle 2$ equations.

$\displaystyle x =\frac{ 7 + \sqrt{4117} }{ 18 }$

$\displaystyle x = 3.95$

$\displaystyle x = - \frac{\sqrt{4117}}{18} + \frac{7}{18}$

$\displaystyle x = -3.18$

Since we want points, we need to plug these values into one of the original equations.

$\displaystyle p_{1} = 9 \cdot 3.95 + 24$

$\displaystyle p_{1} = 35.55 + 24$

$\displaystyle p_{1} = 59.55$

So the first intersection point is at $\displaystyle \left( 3.95 , 59.55 \right)$

Now we need to find the last point of intersection.

$\displaystyle p_{2} = 9 \cdot -3.18 + 24$

$\displaystyle p_{2} = -28.62 + 24$

$\displaystyle p_{2} = -4.62$

So the second intersection point is at $\displaystyle \left( -3.18 , -4.62 \right)$

The first step to solving this problem is to set the equations equal to each other.

$\displaystyle 18 x^{2} - 48 = 10 x + 41$

Now subtract the right hand side, to make it zero, so we can use the quadratic equation.

$\displaystyle 18 x^{2} - 48 - 10 x - 41 = 0$

$\displaystyle 18 x^{2} - 10 x - 89 = 0$

Recall the quadratic equation.

$\displaystyle x =\frac{ -b \pm\sqrt{ b^{2} - 4 a c }}{ 2 a }$

Where $\displaystyle \uptext{a}$, $\displaystyle \uptext{b}$, and $\displaystyle \uptext{c}$ correspond to coefficients in the following quadratic equation.

$\displaystyle a x^{2} + b x + c = 0$

In this case, $\displaystyle a = 18 , b = -10 , c = -89$.

Now plug these values in.

$\displaystyle x =\frac{ 10 \pm\sqrt{\left( -10 \right)^2- 4 \cdot 18 \cdot -89 }}{ 2 \cdot 18 }$

$\displaystyle x =\frac{ 10 \pm 2 \sqrt{1627} }{ 36 }$

Split this up into $\displaystyle 2$ equations.

$\displaystyle x =\frac{ 10 + 2 \sqrt{1627} }{ 36 }$

$\displaystyle x = 2.52$

$\displaystyle x = - \frac{\sqrt{1627}}{18} + \frac{5}{18}$

$\displaystyle x = -1.96$

Since we want points, we need to plug these values into one of the original equations.

$\displaystyle p_{1} = 18 \cdot 2.52 + 41$

$\displaystyle p_{1} = 45.36 + 41$

$\displaystyle p_{1} = 86.36$

So the first intersection point is at $\displaystyle \left( 2.52 , 86.36 \right)$

Now we need to find the last point of intersection.

$\displaystyle p_{2} = 18 \cdot -1.96 + 41$

$\displaystyle p_{2} = -35.28 + 41$

$\displaystyle p_{2} = 5.72$

So the second intersection point is at $\displaystyle \left( -1.96 , 5.72 \right)$

The first step to solving this problem is to set the equations equal to each other.

$\displaystyle 31 x^{2} - 3 = 40 x + 9$

Now subtract the right hand side, to make it zero, so we can use the quadratic equation.

$\displaystyle 31 x^{2} - 3 - 40 x - 9 = 0$

$\displaystyle 31 x^{2} - 40 x - 12 = 0$

Recall the quadratic equation.

$\displaystyle x =\frac{ -b \pm\sqrt{ b^{2} - 4 a c }}{ 2 a }$

Where $\displaystyle \uptext{a}$, $\displaystyle \uptext{b}$, and $\displaystyle \uptext{c}$ correspond to coefficients in the following quadratic equation.

$\displaystyle a x^{2} + b x + c = 0$

In this case,$\displaystyle a = 31 , b = -40 , c = -12$ .

Now plug these values in.

$\displaystyle x =\frac{ 40 \pm\sqrt{\left( -40 \right)^2- 4 \cdot 31 \cdot -12 }}{ 2 \cdot 31 }$

$\displaystyle x =\frac{ 40 \pm 4 \sqrt{193} }{ 62 }$

Split this up into $\displaystyle 2$ equations.

$\displaystyle x =\frac{ 40 + 4 \sqrt{193} }{ 62 }$

$\displaystyle x = 1.54$

$\displaystyle x = - \frac{2 \sqrt{193}}{31} + \frac{20}{31}$

$\displaystyle x = -0.25$

Since we want points, we need to plug these values into one of the original equations.

$\displaystyle p_{1} = 31 \cdot 1.54 + 9$

$\displaystyle p_{1} = 47.74 + 9$

$\displaystyle p_{1} = 56.74$

So the first intersection point is at $\displaystyle \left( 1.54 , 56.74 \right)$

Now we need to find the last point of intersection.

$\displaystyle p_{2} = 31 \cdot -0.25 + 9$

$\displaystyle p_{2} = -7.75 + 9$

$\displaystyle p_{2} = 1.25$

So the second intersection point is at $\displaystyle \left( -0.25 , 1.25 \right)$

The first step to solving this problem is to set the equations equal to each other.

4$\displaystyle x^{2} - 66 = 30 x + 59$

Now subtract the right hand side, to make it zero, so we can use the quadratic equation.

$\displaystyle 4 x^{2} - 66 - 30 x - 59 = 0$

$\displaystyle 4 x^{2} - 30 x - 125 = 0$

Recall the quadratic equation.

$\displaystyle x =\frac{ -b \pm\sqrt{ b^{2} - 4 a c }}{ 2 a }$

Where $\displaystyle \uptext{a}$, $\displaystyle \uptext{b}$, and $\displaystyle \uptext{c}$ correspond to coefficients in the following quadratic equation.

$\displaystyle a x^{2} + b x + c = 0$

In this case, $\displaystyle a = 4 , b = -30 , c = -125$.

Now plug these values in.

$\displaystyle x =\frac{ 30 \pm\sqrt{\left( -30 \right)^2- 4 \cdot 4 \cdot -125 }}{ 2 \cdot 4 }$

$\displaystyle x =\frac{ 30 \pm 10 \sqrt{29} }{ 8 }$

Split this up into $\displaystyle 2$ equations.

$\displaystyle x =\frac{ 30 + 10 \sqrt{29} }{ 8 }$

$\displaystyle x = 10.48$

$\displaystyle x = - \frac{5 \sqrt{29}}{4} + \frac{15}{4}$b

$\displaystyle x = -2.98$

Since we want points, we need to plug these values into one of the original equations.

$\displaystyle p_{1} = 4 \cdot 10.48 + 59$

$\displaystyle p_{1} = 41.92 + 59$

$\displaystyle p_{1} = 100.92$

So the first intersection point is at $\displaystyle \left( 10.48 , 100.92 \right)$

Now we need to find the last point of intersection.

$\displaystyle p_{2} = 4 \cdot -2.98 + 59$

$\displaystyle p_{2} = -11.92 + 59$

$\displaystyle p_{2} = 47.08$

So the second intersection point is at $\displaystyle \left( -2.98 , 47.08 \right)$

The first step to solving this problem is to set the equations equal to each other.

$\displaystyle 37 x^{2} - 89 = 18 x + 75$

Now subtract the right hand side, to make it zero, so we can use the quadratic equation.

$\displaystyle 37 x^{2} - 89 - 18 x - 75 = 0$

$\displaystyle 37 x^{2} - 18 x - 164 = 0$

Recall the quadratic equation.

$\displaystyle x =\frac{ -b \pm\sqrt{ b^{2} - 4 a c }}{ 2 a }$

Where $\displaystyle \uptext{a}$, $\displaystyle \uptext{b}$, and $\displaystyle \uptext{c}$ correspond to coefficients in the following quadratic equation.

$\displaystyle a x^{2} + b x + c = 0$

In this case, $\displaystyle a = 37 , b = -18 , c = -164$.

Now plug these values in.

$\displaystyle x =\frac{ 18 \pm\sqrt{\left( -18 \right)^2- 4 \cdot 37 \cdot -164 }}{ 2 \cdot 37 }$

$\displaystyle x =\frac{ 18 \pm 2 \sqrt{6149} }{ 74 }$

Split this up into $\displaystyle 2$ equations.

$\displaystyle x =\frac{ 18 + 2 \sqrt{6149} }{ 74 }$

$\displaystyle x = 2.36$

$\displaystyle x = - \frac{\sqrt{6149}}{37} + \frac{9}{37}$

$\displaystyle x = -1.88$

Since we want points, we need to plug these values into one of the original equations.

$\displaystyle p_{1} = 37 \cdot 2.36 + 75$

$\displaystyle p_{1} = 87.32 + 75$

$\displaystyle p_{1} = 162.32$

So the first intersection point is at $\displaystyle \left( 2.36 , 162.32 \right)$

Now we need to find the last point of intersection.

$\displaystyle p_{2} = 37 \cdot -1.88 + 75$

$\displaystyle p_{2} = -69.56 + 75$

$\displaystyle p_{2} = 5.44$

So the second intersection point is at $\displaystyle \left( -1.88 , 5.44 \right)$

The first step to solving this problem is to set the equations equal to each other.

$\displaystyle 23 x^{2} - 65 = 11 x + 55$

Now subtract the right hand side, to make it zero, so we can use the quadratic equation.

$\displaystyle 23 x^{2} - 65 - 11 x - 55 = 0$

$\displaystyle 23 x^{2} - 11 x - 120 = 0$

Recall the quadratic equation.

$\displaystyle x =\frac{ -b \pm\sqrt{ b^{2} - 4 a c }}{ 2 a }$

Where $\displaystyle \uptext{a}$, $\displaystyle \uptext{b}$, and $\displaystyle \uptext{c}$ correspond to coefficients in the following quadratic equation.

$\displaystyle a x^{2} + b x + c = 0$

In this case, $\displaystyle a = 23 , b = -11 , c = -120$.

Now plug these values in.

$\displaystyle x =\frac{ 11 \pm\sqrt{\left( -11 \right)^2- 4 \cdot 23 \cdot -120 }}{ 2 \cdot 23 }$

$\displaystyle x =\frac{ 11 \pm \sqrt{11161} }{ 46 }$

Split this up into $\displaystyle 2$ equations.

$\displaystyle x =\frac{ 11 + \sqrt{11161} }{ 46 }$

$\displaystyle x = 2.54$

$\displaystyle x = - \frac{\sqrt{11161}}{46} + \frac{11}{46}$

$\displaystyle x = -2.06$

Since we want points, we need to plug these values into one of the original equations.

$\displaystyle p_{1} = 23 \cdot 2.54 + 55$

$\displaystyle p_{1} = 58.42 + 55$

$\displaystyle p_{1} = 113.42$

So the first intersection point is at $\displaystyle \left( 2.54 , 113.42 \right)$

Now we need to find the last point of intersection.

$\displaystyle p_{2} = 23 \cdot -2.06 + 55$

$\displaystyle p_{2} = -47.38 + 55$

$\displaystyle p_{2} = 7.62$

So the second intersection point is at $\displaystyle \left( -2.06 , 7.62 \right)$

The first step to solving this problem is to set the equations equal to each other.

$\displaystyle 21 x^{2} - 94 = 18 x + 95$

Now subtract the right hand side, to make it zero, so we can use the quadratic equation.

$\displaystyle 21 x^{2} - 94 - 18 x - 95 = 0$

$\displaystyle 21 x^{2} - 18 x - 189 = 0$

Recall the quadratic equation.

$\displaystyle x =\frac{ -b \pm\sqrt{ b^{2} - 4 a c }}{ 2 a }$

Where $\displaystyle \uptext{a}$, $\displaystyle \uptext{b}$, and $\displaystyle \uptext{c}$ correspond to coefficients in the following quadratic equation.

$\displaystyle a x^{2} + b x + c = 0$

In this case, $\displaystyle a = 21 , b = -18 , c = -189$.

Now plug these values in.

$\displaystyle x =\frac{ 18 \pm\sqrt{\left( -18 \right)^2- 4 \cdot 21 \cdot -189 }}{ 2 \cdot 21 }$

$\displaystyle x =\frac{ 18 \pm 90 \sqrt{2} }{ 42 }$

Split this up into $\displaystyle 2$ equations.

$\displaystyle x =\frac{ 18 + 90 \sqrt{2} }{ 42 }$

$\displaystyle x = 3.46$

$\displaystyle x = - \frac{15 \sqrt{2}}{7} + \frac{3}{7}$

$\displaystyle x = -2.6$

Since we want points, we need to plug these values into one of the original equations.

$\displaystyle p_{1} = 21 \cdot 3.46 + 95$

$\displaystyle p_{1} = 72.66 + 95$

$\displaystyle p_{1} = 167.66$

So the first intersection point is at $\displaystyle \left( 3.46 , 167.66 \right)$

Now we need to find the last point of intersection.

$\displaystyle p_{2} = 21 \cdot -2.6 + 95$

$\displaystyle p_{2} = -54.6 + 95$

$\displaystyle p_{2} = 40.4$

So the second intersection point is at $\displaystyle \left( -2.6 , 40.4 \right)$