40x=x^2+10x+100

Solution for 40x=x^2+10x+100 equation:

40x=x^2+10x+100

We move all terms to the left:

40x-(x^2+10x+100)=0

We get rid of parentheses

-x^2+40x-10x-100=0

We add all the numbers together, and all the variables

-1x^2+30x-100=0

a = -1; b = 30; c = -100;
Δ = b2-4ac
Δ = 302-4·(-1)·(-100)
Δ = 500
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:

$\sqrt{\Delta}=\sqrt{500}=\sqrt{100*5}=\sqrt{100}*\sqrt{5}=10\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-10\sqrt{5}}{2*-1}=\frac{-30-10\sqrt{5}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+10\sqrt{5}}{2*-1}=\frac{-30+10\sqrt{5}}{-2}
$