(x+4)(x+4+3x)=152

Solution for (x+4)(x+4+3x)=152 equation:

(x+4)(x+4+3x)=152

We move all terms to the left:

(x+4)(x+4+3x)-(152)=0

We add all the numbers together, and all the variables

(x+4)(4x+4)-152=0

We multiply parentheses ..

(+4x^2+4x+16x+16)-152=0

We get rid of parentheses

4x^2+4x+16x+16-152=0

We add all the numbers together, and all the variables

4x^2+20x-136=0

a = 4; b = 20; c = -136;
Δ = b2-4ac
Δ = 202-4·4·(-136)
Δ = 2576
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:

$\sqrt{\Delta}=\sqrt{2576}=\sqrt{16*161}=\sqrt{16}*\sqrt{161}=4\sqrt{161}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{161}}{2*4}=\frac{-20-4\sqrt{161}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{161}}{2*4}=\frac{-20+4\sqrt{161}}{8}
$