x^2+40x=10

Solution for x^2+40x=10 equation:

x^2+40x=10

We move all terms to the left:

x^2+40x-(10)=0

a = 1; b = 40; c = -10;
Δ = b2-4ac
Δ = 402-4·1·(-10)
Δ = 1640
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:

$\sqrt{\Delta}=\sqrt{1640}=\sqrt{4*410}=\sqrt{4}*\sqrt{410}=2\sqrt{410}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-2\sqrt{410}}{2*1}=\frac{-40-2\sqrt{410}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+2\sqrt{410}}{2*1}=\frac{-40+2\sqrt{410}}{2}
$