\(\displaystyle \small 4x^2+12x+2=0\)

\(\displaystyle \small 4x^2+12+2-2=0-2\)

\(\displaystyle \small 4x^2+12x=-2\)

To complete the square, we have to add a number that makes the left side of the equation a perfect square. Perfect squares have the formula \(\displaystyle \small \small \small a^2+2ab+b^2\).

In this case, \(\displaystyle \small \small a=2x,\ 2ab=12x\).

\(\displaystyle \small \small 12=2(2x)(b)=4xb; b=3\)

\(\displaystyle \small b^2=3^2=9\)

Add this to both sides:

\(\displaystyle \small 4x^2+12x+9=-2+9\)

\(\displaystyle \small \small 4x^2+12x+9=7\)

\(\displaystyle \small (2x+3)^2=7\)

\(\displaystyle \small \sqrt{(2x+3)^2}=\pm \sqrt7\)

\(\displaystyle \small 2x+3=\pm \sqrt{7}\)

\(\displaystyle \small 2x+3-3=-3\pm \sqrt7\)

\(\displaystyle \small 2x=-3\pm \sqrt7\)

\(\displaystyle \small \frac{2x}{2}=\frac{-3\pm \sqrt7}{2}\)

\(\displaystyle \small x=\frac{-3\pm \sqrt7}{2}\)

\(\displaystyle 4x ^{2} - 20x + 25\) can be demonstrated to be a perfect square polynomial as follows:

\(\displaystyle 4x ^{2} - 20x + 25 = \left ( 2x\right )^{2} - 2 (2x)(5) + 5^{2}\)

It can therefore be factored using the pattern

\(\displaystyle A^{2}-2AB + B^{2} = \left (A - B \right )^{2}\)

with \(\displaystyle A = 2x, B = 5\).

We can rewrite and solve the equation accordingly:

\(\displaystyle 4x ^{2} - 20x + 25 = 0\)

\(\displaystyle (2x-5) ^{2} = 0\)

\(\displaystyle 2x-5 = 0\)

\(\displaystyle 2x = 5\)

\(\displaystyle x = 5 \div 2\)

\(\displaystyle x = 2\frac{1}{2}\)

This is the only solution.

When solving a quadratic equation, it is necessary to write it in standard form first - that is, in the form \(\displaystyle ax^{2} + bx + c = 0\). This equation is not in this form, so we must get it in this form as follows:

\(\displaystyle x (x - 7) = 30\)

\(\displaystyle x \cdot x -x \cdot 7 = 30\)

\(\displaystyle x ^{2} -7x = 30\)

\(\displaystyle x ^{2} -7x - 30 = 0\)

We factor the quadratic expression as 

\(\displaystyle (x+M)(x+N)\)

so that \(\displaystyle MN = -30\) and \(\displaystyle M + N = -7\).

By trial and error, we find that 

\(\displaystyle M = -10, N = 3\), so the equation becomes

\(\displaystyle (x-10)(x+3) = 0\)

Set each linear binomial to 0 and solve separately:

\(\displaystyle x - 10 = 0\)

\(\displaystyle x = 10\)

\(\displaystyle x + 3 = 0\)

\(\displaystyle x = -3\)

The solution set is \(\displaystyle \left \{ -3, 10 \right \}\).

When solving a quadratic equation, it is necessary to write it in standard form first - that is, in the form \(\displaystyle ax^{2} + bx + c = 0\). This equation is not in this form, so we must get it in this form as follows:

\(\displaystyle x^{2} - 12 = 6x+4\)

\(\displaystyle x^{2} - 12- 6x - 4 = 6x+4 - 6x - 4\)

\(\displaystyle x^{2} - 6x - 16 = 0\)

We factor the quadratic expression as 

\(\displaystyle (x+M)(x+N)\)

so that \(\displaystyle MN = -16\) and \(\displaystyle M + N = -6\).

By trial and error, we find that 

\(\displaystyle M = -8, N = 2\), so the equation becomes

\(\displaystyle (x-8)(x+2) = 0\).

Set each linear binomial to 0 and solve separately:

\(\displaystyle x-8=0\)

\(\displaystyle x= 8\)

\(\displaystyle x+2= 0\)

\(\displaystyle x= -2\)

The solutions set is \(\displaystyle \left \{ -2, 8\right \}\)

Solve the equation by using the quadratic formula:

\(\displaystyle x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\)

For this equation, \(\displaystyle a=1, b=5, c=-13\). Plug these values into the quadratic equation and to solve for \(\displaystyle x\).

\(\displaystyle x=\frac{-5\pm\sqrt{5^2-4(1)(-13)}}{2(1)}=\frac{-5\pm\sqrt{77}}{2}\)

\(\displaystyle x=1.9\) and \(\displaystyle -6.9\)

Recall the quadratic equation:

\(\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\)

For the given equation, \(\displaystyle a=1, b=5, c=-7\). Plug these into the equation and solve.

\(\displaystyle x=\frac{-5+\sqrt{5^2-4(1)(-7))}}{2(1)}=\frac{-5+\sqrt{25+28}}{2}=\frac{-5+\sqrt{53}}{2}=1.1\)

and

\(\displaystyle x=\frac{-5-\sqrt{5^2-4(1)(-7))}}{2(1)}=\frac{-5-\sqrt{25+28}}{2}=\frac{-5-\sqrt{53}}{2}=-6.1\)

Solve this equation by using the quadratic equation:

\(\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\)

For the equation \(\displaystyle x^2+4x-10=0\), \(\displaystyle a=1, b=4, c=-10\)

Plug it in to the equation to solve for \(\displaystyle x\).

\(\displaystyle x=\frac{-4\pm \sqrt{4^2-4(1)(-10)}}{2}\)

\(\displaystyle x=\frac{-4\pm \sqrt{16+40}}{2}\)

\(\displaystyle x=\frac{-4\pm \sqrt{56}}{2}\)

\(\displaystyle x=1.74\) and \(\displaystyle x=-5.74\)

We have our quadratic equation in the form \(\displaystyle ax^{2}+bx+c=0\)

The quadratic formula is given as:

\(\displaystyle x=\frac{-b_{-}^{+}{\sqrt{b^{2}-4ac}}}{2a}\)

Using \(\displaystyle 2x^{2}+7x-85=0\)

\(\displaystyle x=\frac{-7_{-}^{+}{\sqrt{(7)^{2}-4(2)(-85)}}}{2(2)}\)

\(\displaystyle x=\frac{-7_{-}^{+}{\sqrt{49+680}}}{4}\)

\(\displaystyle x=\frac{-7_{-}^{+}{\sqrt{729}}}{4}\)

\(\displaystyle x=\frac{-7_{-}^{+}{27}}{4}\)

\(\displaystyle x=\frac{-7+27}{4}\)                        \(\displaystyle x=\frac{-7-27}{4}\)

\(\displaystyle x=\frac{20}{4}\)                                   \(\displaystyle x=\frac{-34}{4}\)

\(\displaystyle \textbf{x=5}\)                                       \(\displaystyle \textbf{x=-8.5}\)

To complete the square, we need to get our variable terms on one side and our constant terms on the other.

1) \(\displaystyle x^{2}+8x-3=0\)

\(\displaystyle x^{2}+8x=3\)

2) To make a perfect square trinomial, we need to take one-half of the x-term and square said term. Add the squared term to both sides.

\(\displaystyle x^{2}+8x+{\color{Blue} 16}=3+{\color{Blue} 16}\)

\(\displaystyle x^{2}+8x+16=19\)

3) We now have a perfect square trinomial on the left side which can be represented as a binomial squared. We should check to make sure.

*\(\displaystyle a^{2}+2ab+b^{2}=(a+b)^{2}\) (standard form)

In our equation:

\(\displaystyle (x)^{2}+8x+(4)^{2}\)

\(\displaystyle a=x\)

\(\displaystyle b=4\)

\(\displaystyle 2ab=2(x)(4)=8x\) (CHECK)

4) Represent the perfect square trinomial as a binomial squared:

\(\displaystyle x^{2}+4x+16=(x+4)^{2}\)

5) Take the square root of both sides:

\(\displaystyle (x+4)^{2}=19\)

\(\displaystyle \sqrt{(x+4)^{2}}=_{-}^{+}\sqrt{19}\)

\(\displaystyle (x+4)=_{-}^{+}\sqrt{19}\)

6) Solve for x

\(\displaystyle x= -4_{-}^{+}\sqrt{19}\)

\(\displaystyle x=-4+\sqrt{19}\) or \(\displaystyle x=-4-\sqrt{19}\)

\(\displaystyle 6x^{2}+16x-70=0\) involves rather large numbers, so the Quadratic Formula is applicable here.

\(\displaystyle x=\frac{-b_{-}^{+}\sqrt{b^{2}-4ac}}{2a}\)

\(\displaystyle x=\frac{-16_{-}^{+}\sqrt{(16)^{2}-4(6)(-70)}}{2(6)}\)

\(\displaystyle x=\frac{-16_{-}^{+}\sqrt{256+1680}}{12}\)

\(\displaystyle x=\frac{-16_{-}^{+}\sqrt{1936}}{12}\)

\(\displaystyle x=\frac{-16_{-}^{+}44}{12}\)

\(\displaystyle x=\frac{-16+44}{12}\)         or         \(\displaystyle x=\frac{-16-44}{12}\)

\(\displaystyle x=\frac{28}{12}\)                                  \(\displaystyle x=\frac{-60}{12}\)

\(\displaystyle x=\frac{\textbf{7}}{\textbf{3}}\)                                    \(\displaystyle x=\textbf{-5}\)