X^2+36x-128=0

Solution for X^2+36x-128=0 equation:

X^2+36X-128=0

a = 1; b = 36; c = -128;
Δ = b2-4ac
Δ = 362-4·1·(-128)
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{-(36)-4\sqrt{113}}{2*1}=\frac{-36-4\sqrt{113}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+4\sqrt{113}}{2*1}=\frac{-36+4\sqrt{113}}{2}
$