# How do you solve x^2 + 18x + 80 = 0 by completing the square?

Jul 10, 2015

I found:
${x}_{1} = - 8$
${x}_{2} = - 10$

#### Explanation:

Start by taking $80$ to the right and get:
${x}^{2} + 18 x = - 80$
now you need a number that multiplied by $2$ gives $18$...could be $9$!
So, add and subtract NOT $9$ but ${9}^{2} = 81$!
${x}^{2} + 18 x \textcolor{red}{+ 81 - 81} = - 80$
take $- 81$ to the right:
${x}^{2} + 18 x + 81 = 81 - 80$
so that you can write:
${\left(x + 9\right)}^{2} = 1$
root square both sides:
$x + 9 = \pm \sqrt{1} = \pm 1$
So you get two solutions:
${x}_{1} = - 8$
${x}_{2} = - 10$

Jul 10, 2015

$x = - 8$
or
$x = - 10$

#### Explanation:

Given : ${x}^{2} + 18 x + 80 = 0$

Complete the square:
$\textcolor{w h i t e}{\text{XXXX}}$${x}^{2} + 18 x + {\left(\frac{18}{2}\right)}^{2} - {\left(\frac{18}{2}\right)}^{2} + 80 = 0$
Simplify:
$\textcolor{w h i t e}{\text{XXXX}}$${x}^{2} + 18 x + {9}^{2} - 81 + 80 = 0$
Write as a squared binomial:
$\textcolor{w h i t e}{\text{XXXX}}$${\left(x + 9\right)}^{2} - 1 = 0$
Move the constant to the right side
$\textcolor{w h i t e}{\text{XXXX}}$${\left(x + 9\right)}^{2} = 1$
Take the square root
$\textcolor{w h i t e}{\text{XXXX}}$$x + 9 = \pm 1$
Move constant to the right side
$\textcolor{w h i t e}{\text{XXXX}}$$x = - 10$ or $x = - 8$