4x^2+26x-36=0

Solution for 4x^2+26x-36=0 equation:

4x^2+26x-36=0

a = 4; b = 26; c = -36;
Δ = b2-4ac
Δ = 262-4·4·(-36)
Δ = 1252
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:

$\sqrt{\Delta}=\sqrt{1252}=\sqrt{4*313}=\sqrt{4}*\sqrt{313}=2\sqrt{313}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-2\sqrt{313}}{2*4}=\frac{-26-2\sqrt{313}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+2\sqrt{313}}{2*4}=\frac{-26+2\sqrt{313}}{8}
$