# How do you use the quadratic formula to solve 2(x-3)^2=-2x+9?

May 14, 2017

$x = \frac{5}{2} \pm \frac{1}{2} \sqrt{7}$

#### Explanation:

We can use the quadratic formula:

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

to solve a quadratic equation of the form:

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

So as we have:

$2 {\left(x - 3\right)}^{2} = - 2 x + 9$

The first thing we should do is expand the expression and rearrange into standard form:

$\therefore 2 \left(x - 3\right) \left(x - 3\right) = - 2 x + 9$
$\therefore 2 \left({x}^{2} - 3 x - 3 x + 9\right) = - 2 x + 9$
$\therefore 2 {x}^{2} - 12 x + 18 = - 2 x + 9$
$\therefore 2 {x}^{2} - 10 x + 9 = 0$

We can now apply the quadratic formula:

$x = \frac{- \left(- 10\right) \pm \sqrt{{\left(- 10\right)}^{2} - 4 \left(2\right) \left(9\right)}}{2 \left(2\right)}$
$\setminus \setminus = \frac{10 \pm \sqrt{100 - 72}}{4}$
$\setminus \setminus = \frac{10 \pm \sqrt{28}}{4}$
$\setminus \setminus = \frac{10 \pm 2 \sqrt{7}}{4}$
$\setminus \setminus = \frac{5}{2} \pm \frac{1}{2} \sqrt{7}$