8x^2+8x+8=16

Solution for 8x^2+8x+8=16 equation:

8x^2+8x+8=16

We move all terms to the left:

8x^2+8x+8-(16)=0

We add all the numbers together, and all the variables

8x^2+8x-8=0

a = 8; b = 8; c = -8;
Δ = b2-4ac
Δ = 82-4·8·(-8)
Δ = 320
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:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{5}}{2*8}=\frac{-8-8\sqrt{5}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{5}}{2*8}=\frac{-8+8\sqrt{5}}{16}
$