2(x+3)(x+3)-4=20

Solution for 2(x+3)(x+3)-4=20 equation:

2(x+3)(x+3)-4=20

We move all terms to the left:

2(x+3)(x+3)-4-(20)=0

We add all the numbers together, and all the variables

2(x+3)(x+3)-24=0

We multiply parentheses ..

2(+x^2+3x+3x+9)-24=0

We multiply parentheses

2x^2+6x+6x+18-24=0

We add all the numbers together, and all the variables

2x^2+12x-6=0

a = 2; b = 12; c = -6;
Δ = b2-4ac
Δ = 122-4·2·(-6)
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8\sqrt{3}}{2*2}=\frac{-12-8\sqrt{3}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8\sqrt{3}}{2*2}=\frac{-12+8\sqrt{3}}{4}
$