x^2+240x-1600=0

Solution for x^2+240x-1600=0 equation:

x^2+240x-1600=0

a = 1; b = 240; c = -1600;
Δ = b2-4ac
Δ = 2402-4·1·(-1600)
Δ = 64000
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:

$\sqrt{\Delta}=\sqrt{64000}=\sqrt{6400*10}=\sqrt{6400}*\sqrt{10}=80\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(240)-80\sqrt{10}}{2*1}=\frac{-240-80\sqrt{10}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(240)+80\sqrt{10}}{2*1}=\frac{-240+80\sqrt{10}}{2}
$