2x^2+4x+4=18^2

Solution for 2x^2+4x+4=18^2 equation:

2x^2+4x+4=18^2

We move all terms to the left:

2x^2+4x+4-(18^2)=0

We add all the numbers together, and all the variables

2x^2+4x-320=0

a = 2; b = 4; c = -320;
Δ = b2-4ac
Δ = 42-4·2·(-320)
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{-(4)-4\sqrt{161}}{2*2}=\frac{-4-4\sqrt{161}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{161}}{2*2}=\frac{-4+4\sqrt{161}}{4}
$