x^2+40x-165=0

Solution for x^2+40x-165=0 equation:

x^2+40x-165=0

a = 1; b = 40; c = -165;
Δ = b2-4ac
Δ = 402-4·1·(-165)
Δ = 2260
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:

$\sqrt{\Delta}=\sqrt{2260}=\sqrt{4*565}=\sqrt{4}*\sqrt{565}=2\sqrt{565}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-2\sqrt{565}}{2*1}=\frac{-40-2\sqrt{565}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+2\sqrt{565}}{2*1}=\frac{-40+2\sqrt{565}}{2}
$