15=x^2+4x

Solution for 15=x^2+4x equation:

15=x^2+4x

We move all terms to the left:

15-(x^2+4x)=0

We get rid of parentheses

-x^2-4x+15=0

We add all the numbers together, and all the variables

-1x^2-4x+15=0

a = -1; b = -4; c = +15;
Δ = b2-4ac
Δ = -42-4·(-1)·15
Δ = 76
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:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{19}}{2*-1}=\frac{4-2\sqrt{19}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{19}}{2*-1}=\frac{4+2\sqrt{19}}{-2}
$