2(w+2)2w=228

Solution for 2(w+2)2w=228 equation:

2(w+2)2w=228

We move all terms to the left:

2(w+2)2w-(228)=0

We multiply parentheses

4w^2+8w-228=0

a = 4; b = 8; c = -228;
Δ = b2-4ac
Δ = 82-4·4·(-228)
Δ = 3712
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:

$\sqrt{\Delta}=\sqrt{3712}=\sqrt{64*58}=\sqrt{64}*\sqrt{58}=8\sqrt{58}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{58}}{2*4}=\frac{-8-8\sqrt{58}}{8}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{58}}{2*4}=\frac{-8+8\sqrt{58}}{8}
$