x^2+240x-800=0

Solution for x^2+240x-800=0 equation:

x^2+240x-800=0

a = 1; b = 240; c = -800;
Δ = b2-4ac
Δ = 2402-4·1·(-800)
Δ = 60800
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:

$\sqrt{\Delta}=\sqrt{60800}=\sqrt{1600*38}=\sqrt{1600}*\sqrt{38}=40\sqrt{38}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(240)-40\sqrt{38}}{2*1}=\frac{-240-40\sqrt{38}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(240)+40\sqrt{38}}{2*1}=\frac{-240+40\sqrt{38}}{2}
$