x^2+40x-360=0

Solution for x^2+40x-360=0 equation:

x^2+40x-360=0

a = 1; b = 40; c = -360;
Δ = b2-4ac
Δ = 402-4·1·(-360)
Δ = 3040
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:

$\sqrt{\Delta}=\sqrt{3040}=\sqrt{16*190}=\sqrt{16}*\sqrt{190}=4\sqrt{190}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{190}}{2*1}=\frac{-40-4\sqrt{190}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{190}}{2*1}=\frac{-40+4\sqrt{190}}{2}
$