x^2+50*x-400=0

Solution for x^2+50*x-400=0 equation:

x^2+50x-400=0

a = 1; b = 50; c = -400;
Δ = b2-4ac
Δ = 502-4·1·(-400)
Δ = 4100
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:

$\sqrt{\Delta}=\sqrt{4100}=\sqrt{100*41}=\sqrt{100}*\sqrt{41}=10\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-10\sqrt{41}}{2*1}=\frac{-50-10\sqrt{41}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+10\sqrt{41}}{2*1}=\frac{-50+10\sqrt{41}}{2}
$