4x^2+20x-88=0

Solution for 4x^2+20x-88=0 equation:

4x^2+20x-88=0

a = 4; b = 20; c = -88;
Δ = b2-4ac
Δ = 202-4·4·(-88)
Δ = 1808
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:

$\sqrt{\Delta}=\sqrt{1808}=\sqrt{16*113}=\sqrt{16}*\sqrt{113}=4\sqrt{113}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{113}}{2*4}=\frac{-20-4\sqrt{113}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{113}}{2*4}=\frac{-20+4\sqrt{113}}{8}
$