0=2(x+3)(x+3)-6

Solution for 0=2(x+3)(x+3)-6 equation:

0=2(x+3)(x+3)-6

We move all terms to the left:

0-(2(x+3)(x+3)-6)=0

We add all the numbers together, and all the variables

-(2(x+3)(x+3)-6)=0

We multiply parentheses ..

-(2(+x^2+3x+3x+9)-6)=0


We calculate terms in parentheses: -(2(+x^2+3x+3x+9)-6), so:

2(+x^2+3x+3x+9)-6

We multiply parentheses

2x^2+6x+6x+18-6

We add all the numbers together, and all the variables

2x^2+12x+12

Back to the equation:

-(2x^2+12x+12)


We get rid of parentheses

-2x^2-12x-12=0

a = -2; b = -12; c = -12;
Δ = b2-4ac
Δ = -122-4·(-2)·(-12)
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{3}}{2*-2}=\frac{12-4\sqrt{3}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{3}}{2*-2}=\frac{12+4\sqrt{3}}{-4}
$