4w2+4w+1=8

Solution for 4w2+4w+1=8 equation:

4w^2+4w+1=8

We move all terms to the left:

4w^2+4w+1-(8)=0

We add all the numbers together, and all the variables

4w^2+4w-7=0

a = 4; b = 4; c = -7;
Δ = b2-4ac
Δ = 42-4·4·(-7)
Δ = 128
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8\sqrt{2}}{2*4}=\frac{-4-8\sqrt{2}}{8}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8\sqrt{2}}{2*4}=\frac{-4+8\sqrt{2}}{8}
$