x^2+60x-500=0

Solution for x^2+60x-500=0 equation:

x^2+60x-500=0

a = 1; b = 60; c = -500;
Δ = b2-4ac
Δ = 602-4·1·(-500)
Δ = 5600
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:

$\sqrt{\Delta}=\sqrt{5600}=\sqrt{400*14}=\sqrt{400}*\sqrt{14}=20\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-20\sqrt{14}}{2*1}=\frac{-60-20\sqrt{14}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+20\sqrt{14}}{2*1}=\frac{-60+20\sqrt{14}}{2}
$