x^2+32x+236=0

Solution for x^2+32x+236=0 equation:

x^2+32x+236=0

a = 1; b = 32; c = +236;
Δ = b2-4ac
Δ = 322-4·1·236
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-4\sqrt{5}}{2*1}=\frac{-32-4\sqrt{5}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+4\sqrt{5}}{2*1}=\frac{-32+4\sqrt{5}}{2}
$