4x^2+24x-492=0

Solution for 4x^2+24x-492=0 equation:

4x^2+24x-492=0

a = 4; b = 24; c = -492;
Δ = b2-4ac
Δ = 242-4·4·(-492)
Δ = 8448
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{8448}=\sqrt{256*33}=\sqrt{256}*\sqrt{33}=16\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-16\sqrt{33}}{2*4}=\frac{-24-16\sqrt{33}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+16\sqrt{33}}{2*4}=\frac{-24+16\sqrt{33}}{8}
$