x^2+25x-2500=0

Solution for x^2+25x-2500=0 equation:

x^2+25x-2500=0

a = 1; b = 25; c = -2500;
Δ = b2-4ac
Δ = 252-4·1·(-2500)
Δ = 10625
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:

$\sqrt{\Delta}=\sqrt{10625}=\sqrt{625*17}=\sqrt{625}*\sqrt{17}=25\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-25\sqrt{17}}{2*1}=\frac{-25-25\sqrt{17}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+25\sqrt{17}}{2*1}=\frac{-25+25\sqrt{17}}{2}
$