text: # 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 ${x}_{1 / 2} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
  $a = 2$
  $b = - 5$
  $c = - 7$
  
  ${x}_{1 / 2} = \frac{5 \pm \sqrt{\left(25 - 4 \cdot 2 \left(- 7\right)\right)}}{2 \cdot 2}$
  
  ${x}_{1 / 2} = \frac{5 \pm \sqrt{\left(25 + 56\right)}}{4}$
  
  ${x}_{1 / 2} = \frac{5 \pm \sqrt{81}}{4}$
  
  ${x}_{1 / 2} = \frac{5 \pm 9}{4}$
  
  ${x}_{1} = \frac{5 + 9}{4} = \frac{14}{4} = \frac{7}{2}$
  
  ${x}_{2} = \frac{5 - 9}{4} = \frac{- 4}{4} = - 1$