# How do you factor (m-7)= -(m-7)^2 + 3?

Oct 29, 2016

$m = \frac{13}{2} \pm \frac{\sqrt{13}}{2}$

or

$\left(m - \frac{13}{2} - \frac{\sqrt{13}}{2}\right) \left(m - \frac{13}{2} + \frac{\sqrt{13}}{2}\right) = 0$

#### Explanation:

Given:

$\left(m - 7\right) = - {\left(m - 7\right)}^{2} + 3$

First add ${\left(m - 7\right)}^{2} - 3$ to both sides to get:

${\left(m - 7\right)}^{2} + \left(m - 7\right) - 3 = 0$

This is in the form:

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

with $x = \left(m - 7\right)$, $a = \textcolor{p u r p \le}{1}$, $b = \textcolor{b l u e}{1}$ and $c = \textcolor{b r o w n}{- 3}$

So we can use the quadratic formula to find:

$m - 7 = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$\textcolor{w h i t e}{m - 7} = \frac{- \textcolor{b l u e}{1} \pm \sqrt{{\textcolor{b l u e}{1}}^{2} - 4 \left(\textcolor{p u r p \le}{1}\right) \left(\textcolor{b r o w n}{- 3}\right)}}{2 \left(\textcolor{p u r p \le}{1}\right)}$

$\textcolor{w h i t e}{m - 7} = \frac{- 1 \pm \sqrt{1 + 12}}{2}$

$\textcolor{w h i t e}{m - 7} = - \frac{1}{2} \pm \frac{\sqrt{13}}{2}$

Add $7$ to both ends to find:

$m = \frac{13}{2} \pm \frac{\sqrt{13}}{2}$