4x^2+36x-52=0

Solution for 4x^2+36x-52=0 equation:

4x^2+36x-52=0

a = 4; b = 36; c = -52;
Δ = b2-4ac
Δ = 362-4·4·(-52)
Δ = 2128
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:

$\sqrt{\Delta}=\sqrt{2128}=\sqrt{16*133}=\sqrt{16}*\sqrt{133}=4\sqrt{133}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-4\sqrt{133}}{2*4}=\frac{-36-4\sqrt{133}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+4\sqrt{133}}{2*4}=\frac{-36+4\sqrt{133}}{8}
$