15^2=x^2+40x

Solution for 15^2=x^2+40x equation:

15^2=x^2+40x

We move all terms to the left:

15^2-(x^2+40x)=0

We add all the numbers together, and all the variables

-(x^2+40x)+225=0

We get rid of parentheses

-x^2-40x+225=0

We add all the numbers together, and all the variables

-1x^2-40x+225=0

a = -1; b = -40; c = +225;
Δ = b2-4ac
Δ = -402-4·(-1)·225
Δ = 2500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2500}=50$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-50}{2*-1}=\frac{-10}{-2}
=+5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+50}{2*-1}=\frac{90}{-2}
=-45
$