# How do you solve x + 9 = 2(x - 1)^2?

Apr 15, 2018

${x}_{1} = \frac{7}{2}$
${x}_{2} = - 1$

#### Explanation:

$x + 9 = 2 \left({x}^{2} - 2 x + 1\right)$

$x + 9 = 2 {x}^{2} - 4 x + 2$

$- 2 {x}^{2} + x + 4 x + 9 - 2 = 0$

$- 2 {x}^{2} + 5 x + 7 = 0 / \cdot \left(- 1\right)$

$2 {x}^{2} - 5 x - 7 = 0$

Now use the formula ${x}_{1 / 2} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$a = 2$
$b = - 5$
$c = - 7$

${x}_{1 / 2} = \frac{5 \pm \sqrt{\left(25 - 4 \cdot 2 \left(- 7\right)\right)}}{2 \cdot 2}$

${x}_{1 / 2} = \frac{5 \pm \sqrt{\left(25 + 56\right)}}{4}$

${x}_{1 / 2} = \frac{5 \pm \sqrt{81}}{4}$

${x}_{1 / 2} = \frac{5 \pm 9}{4}$

${x}_{1} = \frac{5 + 9}{4} = \frac{14}{4} = \frac{7}{2}$

${x}_{2} = \frac{5 - 9}{4} = \frac{- 4}{4} = - 1$