2x+40=x^2+20

Solution for 2x+40=x^2+20 equation:

2x+40=x^2+20

We move all terms to the left:

2x+40-(x^2+20)=0

We get rid of parentheses

-x^2+2x-20+40=0

We add all the numbers together, and all the variables

-1x^2+2x+20=0

a = -1; b = 2; c = +20;
Δ = b2-4ac
Δ = 22-4·(-1)·20
Δ = 84
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:

$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{21}}{2*-1}=\frac{-2-2\sqrt{21}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{21}}{2*-1}=\frac{-2+2\sqrt{21}}{-2}
$