# How do you find the roots, real and imaginary, of y= -3x^2 -5x +3(x/2+1)^2  using the quadratic formula?

Feb 23, 2018

$\frac{- 4 \pm 2 \sqrt{31}}{9}$

#### Explanation:

To solve for roots, we have to substitute y as 0.

$- 3 {x}^{2} - 5 x + 3 {\left(\frac{x}{2} + 1\right)}^{2} = 0$
$- 3 {x}^{2} - 5 x + 3 \left({x}^{2} / 4 + x + 1\right) = 0$
$- 3 {x}^{2} - 5 x + \frac{3 {x}^{2}}{4} + 3 x + 3 = 0$
$- 12 {x}^{2} - 20 x + 3 {x}^{2} + 12 x + 12 = 0$
$- 9 {x}^{2} - 8 x + 12 = 0$

$9 {x}^{2} + 8 x - 12 = 0$

First, let confirm whether the roots are real or imaginary.
$\delta = {8}^{2} - 4 \left(9\right) \left(- 12\right) = 496 > 0$
Hence, both roots are real.

For a quadratic equation $A {x}^{2} + B x + C = 0$, the roots are
$\frac{- B \pm \sqrt{{B}^{2} - 4 A C}}{2 A}$

Therefore the req. roots
$= \frac{- 8 \pm \sqrt{\delta}}{2 \cdot 9}$

$= \frac{- 8 \pm 4 \sqrt{31}}{18}$

$= \frac{- 4 \pm 2 \sqrt{31}}{9}$

Note:
1. $\delta$ refers to the value in the square root in the quadratic equation.
2. $\pm$ means that $+$ and $-$ can contribute to two equally correct answers, in this case, the two roots are actually $\frac{- 4 + 2 \sqrt{31}}{9}$ and $\frac{- 4 - 2 \sqrt{31}}{9}$