# How do you use the discriminant to determine the nature of the roots for -6x^2 - 12x + 90 = 0?

Jun 19, 2015

As here, $\Delta > 0$ and a perfect square there are two real rational roots.

#### Explanation:

$- 6 {x}^{2} - 12 x + 90 = 0$
The equation is of the form color(blue)(ax^2+bx+c=0 where:
$a = - 6 , b = - 12 , c = 90$

The Discriminant is given by:
$\Delta = {b}^{2} - 4 \cdot a \cdot c$
$= {\left(- 12\right)}^{2} - \left(4 \cdot \left(- 6\right) \cdot 90\right)$
$= 144 + 2160 = 2304$

As here, $\Delta > 0$ and it is also a perfect square there are two real rational roots.

• Note :
The solutions are normally found using the formula
$x = \frac{- b \pm \sqrt{\Delta}}{2 \cdot a}$

As $\Delta = 2304$, $x = \frac{- \left(- 12\right) \pm \sqrt{2304}}{2 \cdot - 6} = \frac{12 \pm 48}{-} 12$
 x = (12-48)/-12 = 36/12= color(green)(3
 x = (12+48)/-12 = -60/12 =color(green)( -5