# How do you find the roots, real and imaginary, of y= x^2 - 8x -9- (2x-1)^2  using the quadratic formula?

Dec 22, 2017

see below

#### Explanation:

x^2−8x−9−(2x−1)^2

=>x^2−8x−9−(4x^2-2x+1)

=>x^2−8x−9−4x^2+2x-1)

=>-3x^2−6x−10

$\implies 3 {x}^{2} + 6 x + 10$

$D = {b}^{2} - 4 a c \quad \implies \quad 36 - 4 \cdot 3 \cdot 10$
$D < 0 = 36 - 120 = - 84$

${x}_{1 , 2} = \frac{- b \pm \sqrt{D}}{2 a} = \frac{- 6 \pm \sqrt{7 \cdot 3 \cdot 2 \cdot 2 \cdot {i}^{2}}}{6}$

${x}_{1 , 2} = \frac{- 6 \pm 2 i \sqrt{7 \cdot 3}}{6} = \frac{\cancel{2} \left(- 3 \pm i \sqrt{21}\right)}{\cancel{2} \cdot 3}$

${x}_{1} = \frac{- 3 - i \sqrt{21}}{3} = - 1 - \frac{i \sqrt{21}}{3}$

${x}_{2} = \frac{- 3 + i \sqrt{21}}{3} = - 1 + \frac{i \sqrt{21}}{3}$