x^2+50x-2400=0

Solution for x^2+50x-2400=0 equation:

x^2+50x-2400=0

a = 1; b = 50; c = -2400;
Δ = b2-4ac
Δ = 502-4·1·(-2400)
Δ = 12100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12100}=110$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-110}{2*1}=\frac{-160}{2}
=-80
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+110}{2*1}=\frac{60}{2}
=30
$