4x^2+160x-156=0

Solution for 4x^2+160x-156=0 equation:

4x^2+160x-156=0

a = 4; b = 160; c = -156;
Δ = b2-4ac
Δ = 1602-4·4·(-156)
Δ = 28096
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:

$\sqrt{\Delta}=\sqrt{28096}=\sqrt{64*439}=\sqrt{64}*\sqrt{439}=8\sqrt{439}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(160)-8\sqrt{439}}{2*4}=\frac{-160-8\sqrt{439}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(160)+8\sqrt{439}}{2*4}=\frac{-160+8\sqrt{439}}{8}
$