40x+x^2=100

Solution for 40x+x^2=100 equation:

40x+x^2=100

We move all terms to the left:

40x+x^2-(100)=0

a = 1; b = 40; c = -100;
Δ = b2-4ac
Δ = 402-4·1·(-100)
Δ = 2000
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:

$\sqrt{\Delta}=\sqrt{2000}=\sqrt{400*5}=\sqrt{400}*\sqrt{5}=20\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-20\sqrt{5}}{2*1}=\frac{-40-20\sqrt{5}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+20\sqrt{5}}{2*1}=\frac{-40+20\sqrt{5}}{2}
$