4x^2-16x-96=0

Solution for 4x^2-16x-96=0 equation:

4x^2-16x-96=0

a = 4; b = -16; c = -96;
Δ = b2-4ac
Δ = -162-4·4·(-96)
Δ = 1792
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:

$\sqrt{\Delta}=\sqrt{1792}=\sqrt{256*7}=\sqrt{256}*\sqrt{7}=16\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16\sqrt{7}}{2*4}=\frac{16-16\sqrt{7}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16\sqrt{7}}{2*4}=\frac{16+16\sqrt{7}}{8}
$