4x^2+80x-96=0

Solution for 4x^2+80x-96=0 equation:

4x^2+80x-96=0

a = 4; b = 80; c = -96;
Δ = b2-4ac
Δ = 802-4·4·(-96)
Δ = 7936
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:

$\sqrt{\Delta}=\sqrt{7936}=\sqrt{256*31}=\sqrt{256}*\sqrt{31}=16\sqrt{31}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-16\sqrt{31}}{2*4}=\frac{-80-16\sqrt{31}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+16\sqrt{31}}{2*4}=\frac{-80+16\sqrt{31}}{8}
$