X^2+20x-22500=0

Solution for X^2+20x-22500=0 equation:

X^2+20X-22500=0

a = 1; b = 20; c = -22500;
Δ = b2-4ac
Δ = 202-4·1·(-22500)
Δ = 90400
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:

$\sqrt{\Delta}=\sqrt{90400}=\sqrt{400*226}=\sqrt{400}*\sqrt{226}=20\sqrt{226}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20\sqrt{226}}{2*1}=\frac{-20-20\sqrt{226}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20\sqrt{226}}{2*1}=\frac{-20+20\sqrt{226}}{2}
$