x^2+20x=2250

Solution for x^2+20x=2250 equation:

x^2+20x=2250

We move all terms to the left:

x^2+20x-(2250)=0

a = 1; b = 20; c = -2250;
Δ = b2-4ac
Δ = 202-4·1·(-2250)
Δ = 9400
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:

$\sqrt{\Delta}=\sqrt{9400}=\sqrt{100*94}=\sqrt{100}*\sqrt{94}=10\sqrt{94}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-10\sqrt{94}}{2*1}=\frac{-20-10\sqrt{94}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+10\sqrt{94}}{2*1}=\frac{-20+10\sqrt{94}}{2}
$