# How do you solve y=5x^2+20x+23 using the completing square method?

Dec 27, 2016

$x = - 2 \pm \sqrt{\frac{y - 3}{5}}$

#### Explanation:

I will assume that we want to solve the equation in the sense of expressing possible values of $x$ in terms of $y$.

If you actually just wanted to know the zeros, we can then put $y = 0$ and evaluate the result.

Given:

$y = 5 {x}^{2} + 20 x + 23$

We will perform a sequence of steps to isolate $x$ on one side.

First divide both sides by $5$ to make the coefficient of ${x}^{2}$ into $1$:

$\frac{y}{5} = {x}^{2} + 4 x + \frac{23}{5}$

If we take half of the coefficient of $x$ then we get the value $2$, so note that:

${\left(x + 2\right)}^{2} = {x}^{2} + 4 x + 4$

which matches the right hand side of our equation in the first two terms.

So we can proceed as follows:

$\frac{y}{5} = {x}^{2} + 4 x + \frac{23}{5}$

$\textcolor{w h i t e}{\frac{y}{5}} = {x}^{2} + 4 x + 4 - 4 + \frac{23}{5}$

$\textcolor{w h i t e}{\frac{y}{5}} = {\left(x + 2\right)}^{2} + \frac{3}{5}$

Subtracting $\frac{3}{5}$ from both ends we get:

$\frac{y - 3}{5} = {\left(x + 2\right)}^{2}$

Transposed:

${\left(x + 2\right)}^{2} = \frac{y - 3}{5}$

Take the square root of both sides, allowing for the possibility of either sign of square root to get:

$x + 2 = \pm \sqrt{\frac{y - 3}{5}}$

Finally subtract $2$ from both sides to find:

$x = - 2 \pm \sqrt{\frac{y - 3}{5}}$

If we put $y = 0$ to find the zero of the quadratic then we get a pair of non-Real Complex values:

$x = - 2 \pm \sqrt{\frac{0 - 3}{5}} = - 2 \pm \sqrt{- \frac{3}{5}} = - 2 \pm \sqrt{\frac{3}{5}} i = - 2 \pm \frac{\sqrt{15}}{5} i$