# How do you find the roots, real and imaginary, of y=-2x^2 + 7x +12(x/2-1)^2  using the quadratic formula?

Nov 26, 2017

${x}_{1} = \frac{5}{2} + \frac{\sqrt{23}}{2} i \approx 2.5 + 2.4 i$
${x}_{2} = \frac{5}{2} - \frac{\sqrt{23}}{2} i \approx 2.5 - 2.4 i$

When $\frac{5}{2}$ in real and $\frac{\sqrt{23}}{2}$ is imaginary for both

#### Explanation:

for the form of:
$y = a {x}^{2} + b x + c$

${x}_{1 , 2} = \frac{- b \pm \sqrt{{b}^{2} - 4 \cdot a \cdot c}}{2 \cdot a}$

$y = - 2 {x}^{2} + 7 x + 12 {\left(\frac{x}{2} - 1\right)}^{2}$

$\implies y = - 2 {x}^{2} + 7 x + 12 \left({x}^{2} / 4 + 2 \cdot \left(\frac{x}{2}\right) \cdot \left(- 1\right) + 1\right)$

$\implies y = - 2 {x}^{2} + 7 x + 12 \left({x}^{2} / 4 - x + 1\right)$

$\implies y = - 2 {x}^{2} + 7 x + 3 {x}^{2} - 12 x + 12$

$\implies y = {x}^{2} - 5 x + 12$

$\implies$
$a = 1$
$b = - 5$
$c = 12$

$\implies$
${x}_{1 , 2} = \frac{- b \pm \sqrt{{b}^{2} - 4 \cdot a \cdot c}}{2 \cdot a} = \frac{- \left(- 5\right) \pm \sqrt{25 - 4 \cdot 1 \cdot 12}}{2 \cdot 1} =$
$= \frac{5 \pm \sqrt{25 - 48}}{2} =$
$= \frac{5 \pm \sqrt{- 23}}{2} =$
$= \frac{5 \pm \sqrt{23} i}{2} =$
$\implies$
${x}_{1} = \frac{5 + \sqrt{23} i}{2} = \frac{5}{2} + \frac{\sqrt{23}}{2} i = 2.5 + \frac{\sqrt{23}}{2} i \approx 2.5 + 2.4 i$
${x}_{2} = \frac{5 - \sqrt{23} i}{2} = \frac{5}{2} - \frac{\sqrt{23}}{2} i = 2.5 - \frac{\sqrt{23}}{2} i \approx 2.5 - 2.4 i$