x^2+40x-256=0

Solution for x^2+40x-256=0 equation:

x^2+40x-256=0

a = 1; b = 40; c = -256;
Δ = b2-4ac
Δ = 402-4·1·(-256)
Δ = 2624
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:

$\sqrt{\Delta}=\sqrt{2624}=\sqrt{64*41}=\sqrt{64}*\sqrt{41}=8\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8\sqrt{41}}{2*1}=\frac{-40-8\sqrt{41}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8\sqrt{41}}{2*1}=\frac{-40+8\sqrt{41}}{2}
$