4x+16=x^2+2x

Solution for 4x+16=x^2+2x equation:

4x+16=x^2+2x

We move all terms to the left:

4x+16-(x^2+2x)=0

We get rid of parentheses

-x^2+4x-2x+16=0

We add all the numbers together, and all the variables

-1x^2+2x+16=0

a = -1; b = 2; c = +16;
Δ = b2-4ac
Δ = 22-4·(-1)·16
Δ = 68
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:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{17}}{2*-1}=\frac{-2-2\sqrt{17}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{17}}{2*-1}=\frac{-2+2\sqrt{17}}{-2}
$