x^2+2x=x+40

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

x^2+2x=x+40

We move all terms to the left:

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

We get rid of parentheses

x^2+2x-x-40=0

We add all the numbers together, and all the variables

x^2+x-40=0

a = 1; b = 1; c = -40;
Δ = b2-4ac
Δ = 12-4·1·(-40)
Δ = 161
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{161}}{2*1}=\frac{-1-\sqrt{161}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{161}}{2*1}=\frac{-1+\sqrt{161}}{2}
$