x^2+40x=125

Solution for x^2+40x=125 equation:

x^2+40x=125

We move all terms to the left:

x^2+40x-(125)=0

a = 1; b = 40; c = -125;
Δ = b2-4ac
Δ = 402-4·1·(-125)
Δ = 2100
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:

$\sqrt{\Delta}=\sqrt{2100}=\sqrt{100*21}=\sqrt{100}*\sqrt{21}=10\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-10\sqrt{21}}{2*1}=\frac{-40-10\sqrt{21}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+10\sqrt{21}}{2*1}=\frac{-40+10\sqrt{21}}{2}
$