x^2+(2x+1)=16

Solution for x^2+(2x+1)=16 equation:

x^2+(2x+1)=16

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2+2x-15=0

a = 1; b = 2; c = -15;
Δ = b2-4ac
Δ = 22-4·1·(-15)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-8}{2*1}=\frac{-10}{2}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+8}{2*1}=\frac{6}{2}
=3
$