8=w^2-4w

Solution for 8=w^2-4w equation:

8=w^2-4w

We move all terms to the left:

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

We get rid of parentheses

-w^2+4w+8=0

We add all the numbers together, and all the variables

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

a = -1; b = 4; c = +8;
Δ = b2-4ac
Δ = 42-4·(-1)·8
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{3}}{2*-1}=\frac{-4-4\sqrt{3}}{-2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{3}}{2*-1}=\frac{-4+4\sqrt{3}}{-2}
$