x^2+50x-408=0

Solution for x^2+50x-408=0 equation:

x^2+50x-408=0

a = 1; b = 50; c = -408;
Δ = b2-4ac
Δ = 502-4·1·(-408)
Δ = 4132
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:

$\sqrt{\Delta}=\sqrt{4132}=\sqrt{4*1033}=\sqrt{4}*\sqrt{1033}=2\sqrt{1033}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-2\sqrt{1033}}{2*1}=\frac{-50-2\sqrt{1033}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+2\sqrt{1033}}{2*1}=\frac{-50+2\sqrt{1033}}{2}
$