2(x^2+4x+4)-5=8

Solution for 2(x^2+4x+4)-5=8 equation:

2(x^2+4x+4)-5=8

We move all terms to the left:

2(x^2+4x+4)-5-(8)=0

We add all the numbers together, and all the variables

2(x^2+4x+4)-13=0

We multiply parentheses

2x^2+8x+8-13=0

We add all the numbers together, and all the variables

2x^2+8x-5=0

a = 2; b = 8; c = -5;
Δ = b2-4ac
Δ = 82-4·2·(-5)
Δ = 104
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:

$\sqrt{\Delta}=\sqrt{104}=\sqrt{4*26}=\sqrt{4}*\sqrt{26}=2\sqrt{26}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{26}}{2*2}=\frac{-8-2\sqrt{26}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{26}}{2*2}=\frac{-8+2\sqrt{26}}{4}
$