2x^2+32x+36=0

Solution for 2x^2+32x+36=0 equation:

2x^2+32x+36=0

a = 2; b = 32; c = +36;
Δ = b2-4ac
Δ = 322-4·2·36
Δ = 736
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:

$\sqrt{\Delta}=\sqrt{736}=\sqrt{16*46}=\sqrt{16}*\sqrt{46}=4\sqrt{46}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-4\sqrt{46}}{2*2}=\frac{-32-4\sqrt{46}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+4\sqrt{46}}{2*2}=\frac{-32+4\sqrt{46}}{4}
$