x^2+20x=1925

Solution for x^2+20x=1925 equation:

x^2+20x=1925

We move all terms to the left:

x^2+20x-(1925)=0

a = 1; b = 20; c = -1925;
Δ = b2-4ac
Δ = 202-4·1·(-1925)
Δ = 8100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{8100}=90$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-90}{2*1}=\frac{-110}{2}
=-55
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+90}{2*1}=\frac{70}{2}
=35
$