(2x+1)(2x+1)+38=180

Solution for (2x+1)(2x+1)+38=180 equation:

(2x+1)(2x+1)+38=180

We move all terms to the left:

(2x+1)(2x+1)+38-(180)=0

We add all the numbers together, and all the variables

(2x+1)(2x+1)-142=0

We multiply parentheses ..

(+4x^2+2x+2x+1)-142=0

We get rid of parentheses

4x^2+2x+2x+1-142=0

We add all the numbers together, and all the variables

4x^2+4x-141=0

a = 4; b = 4; c = -141;
Δ = b2-4ac
Δ = 42-4·4·(-141)
Δ = 2272
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:

$\sqrt{\Delta}=\sqrt{2272}=\sqrt{16*142}=\sqrt{16}*\sqrt{142}=4\sqrt{142}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{142}}{2*4}=\frac{-4-4\sqrt{142}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{142}}{2*4}=\frac{-4+4\sqrt{142}}{8}
$