# How do you find the roots, real and imaginary, of y=-(x+5)^2+2x^2 + 5x - 12  using the quadratic formula?

Jul 11, 2017

${x}_{1} \approx 6.437 + 0 i$
${x}_{2} \approx - 1.437 + 0 i$

#### Explanation:

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

Before we proceed, let's us expand the equation of $y$
$y = - {\left(x + 5\right)}^{2} + 2 {x}^{2} + 5 x - 12$
$y = - \left({x}^{2} + 10 x + 25\right) + 2 {x}^{2} + 5 x - 12$
$y = - {x}^{2} - 10 x - 25 + 2 {x}^{2} + 5 x - 12$
$y = \left(2 - 1\right) {x}^{2} + \left(5 - 10\right) x + \left(- 37\right)$
$y = \left(1\right) {x}^{2} + \left(- 5\right) x + \left(- 37\right)$

Determine the $a$,$b$, and $c$
$a = 1$
$b = - 5$
$c = - 37$

Use the formula:
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$x = \frac{- \left(- 5\right) \pm \sqrt{{\left(- 5\right)}^{2} - 4 \left(1\right) \left(- 37\right)}}{2 \left(1\right)}$
$x = \frac{5 \pm \sqrt{25 + 37}}{2}$
$x = \frac{5 \pm \sqrt{62}}{2}$

${x}_{1} = \frac{5 + \sqrt{62}}{2} \approx 6.437$
${x}_{2} = \frac{5 - \sqrt{62}}{2} \approx - 1.437$

In this case, the roots are real, their imaginary parts are 0
${x}_{1} \approx 6.437 + 0 i$
${x}_{2} \approx - 1.437 + 0 i$