# How do you find the roots, real and imaginary, of y=-3x^2 + 12x +14  using the quadratic formula?

Aug 1, 2018

$x = - 0.94$ and $x = 4.94$

#### Explanation:

The general form of a quadratic equation is $y = a {x}^{2} + b x + c$

The quadratic formula is $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ where $a , b , c$ are from the general form

Looking at $y = - 3 {x}^{2} + 12 x + 14$,
$a = - 3$
$b = 12$
$c = 14$

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

$x = \frac{- 12 \pm \sqrt{144 + 168}}{2 \times - 3}$

$x = \frac{- 12 \pm \sqrt{312}}{-} 6$

$x = \frac{- 12 + \sqrt{312}}{-} 6$ or $x = \frac{- 12 - \sqrt{312}}{-} 6$

Therefore, the roots are $x = - 0.94$ and $x = 4.94$