x^2+40x-680=0

Solution for x^2+40x-680=0 equation:

x^2+40x-680=0

a = 1; b = 40; c = -680;
Δ = b2-4ac
Δ = 402-4·1·(-680)
Δ = 4320
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:

$\sqrt{\Delta}=\sqrt{4320}=\sqrt{144*30}=\sqrt{144}*\sqrt{30}=12\sqrt{30}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-12\sqrt{30}}{2*1}=\frac{-40-12\sqrt{30}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+12\sqrt{30}}{2*1}=\frac{-40+12\sqrt{30}}{2}
$