x^2+40x+320=0

Solution for x^2+40x+320=0 equation:

x^2+40x+320=0

a = 1; b = 40; c = +320;
Δ = b2-4ac
Δ = 402-4·1·320
Δ = 320
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:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8\sqrt{5}}{2*1}=\frac{-40-8\sqrt{5}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8\sqrt{5}}{2*1}=\frac{-40+8\sqrt{5}}{2}
$