2x^2+4x+4=5

Solution for 2x^2+4x+4=5 equation:

2x^2+4x+4=5

We move all terms to the left:

2x^2+4x+4-(5)=0

We add all the numbers together, and all the variables

2x^2+4x-1=0

a = 2; b = 4; c = -1;
Δ = b2-4ac
Δ = 42-4·2·(-1)
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{6}}{2*2}=\frac{-4-2\sqrt{6}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{6}}{2*2}=\frac{-4+2\sqrt{6}}{4}
$