# How do you find the roots, real and imaginary, of y=-5x^2 + 17x +12(x/2-1)^2  using the quadratic formula?

Jan 23, 2018

$x = 4$

or
$x = \frac{- 3}{2}$

#### Explanation:

$- 5 {x}^{2} + 17 x + 12 {\left(\frac{x}{2} - 1\right)}^{2} = 0$

First we simplify the equation to bring it to the standard form.
Therefore we need to simplify the third term of the equation using ${\left(a - b\right)}^{2} = {a}^{2} + {b}^{2} - 2 a b$

$- 5 {x}^{2} + 17 x + 12 \left({\left(\frac{x}{2}\right)}^{2} + {1}^{2} - 2 \cdot \frac{x}{2} \cdot 1\right) = 0$

$- 5 {x}^{2} + 17 x + 12 \left({x}^{2} / 4 + 1 - x\right) = 0$

$- 5 {x}^{2} + 17 x + \frac{12 {x}^{2}}{4} + 12 - 12 x = 0$

$- 5 {x}^{2} + 17 x + 3 {x}^{2} + 12 - 12 x = 0$

$- 2 {x}^{2} + 5 x + 12 = 0$

$2 {x}^{2} - 5 x - 12 = 0$

Here $a = 2 , b = \left(- 5\right) , c = \left(- 12\right)$

This is the standard form of the equation.
Now we can simply use the quadratic formula to find the roots of this equation.

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

$x = \frac{5 \pm \sqrt{25 + 96}}{4}$

$x = \frac{5 \pm 11}{4}$

$x = \frac{5 + 11}{4} , \frac{5 - 11}{4}$

$x = 4$ and $x = \frac{- 3}{2}$

The Delta ($\sqrt{{b}^{2} - 4 a c}$) of the equation was greater than zero therefore the roots are real and distinct.
If the Delta was equal to zero, the roots would've been real and equal.
If the Delta was less than zero the roots would've been imaginary and imaginary roots always come in pairs.