15=x^2+3+4+2x

Solution for 15=x^2+3+4+2x equation:

15=x^2+3+4+2x

We move all terms to the left:

15-(x^2+3+4+2x)=0

We get rid of parentheses

-x^2-2x-3-4+15=0

We add all the numbers together, and all the variables

-1x^2-2x+8=0

a = -1; b = -2; c = +8;
Δ = b2-4ac
Δ = -22-4·(-1)·8
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-6}{2*-1}=\frac{-4}{-2}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+6}{2*-1}=\frac{8}{-2}
=-4
$