4x^2-22x-96=0.

Solution for 4x^2-22x-96=0. equation:

4x^2-22x-96=0.

a = 4; b = -22; c = -96;
Δ = b2-4ac
Δ = -222-4·4·(-96)
Δ = 2020
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:

$\sqrt{\Delta}=\sqrt{2020}=\sqrt{4*505}=\sqrt{4}*\sqrt{505}=2\sqrt{505}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{505}}{2*4}=\frac{22-2\sqrt{505}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{505}}{2*4}=\frac{22+2\sqrt{505}}{8}
$