(2x+4)(2x+4)=48

Solution for (2x+4)(2x+4)=48 equation:

(2x+4)(2x+4)=48

We move all terms to the left:

(2x+4)(2x+4)-(48)=0

We multiply parentheses ..

(+4x^2+8x+8x+16)-48=0

We get rid of parentheses

4x^2+8x+8x+16-48=0

We add all the numbers together, and all the variables

4x^2+16x-32=0

a = 4; b = 16; c = -32;
Δ = b2-4ac
Δ = 162-4·4·(-32)
Δ = 768
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:

$\sqrt{\Delta}=\sqrt{768}=\sqrt{256*3}=\sqrt{256}*\sqrt{3}=16\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16\sqrt{3}}{2*4}=\frac{-16-16\sqrt{3}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16\sqrt{3}}{2*4}=\frac{-16+16\sqrt{3}}{8}
$