w2+4w+12=228

Solution for w2+4w+12=228 equation:

w2+4w+12=228

We move all terms to the left:

w2+4w+12-(228)=0

We add all the numbers together, and all the variables

w^2+4w-216=0

a = 1; b = 4; c = -216;
Δ = b2-4ac
Δ = 42-4·1·(-216)
Δ = 880
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:

$\sqrt{\Delta}=\sqrt{880}=\sqrt{16*55}=\sqrt{16}*\sqrt{55}=4\sqrt{55}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{55}}{2*1}=\frac{-4-4\sqrt{55}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{55}}{2*1}=\frac{-4+4\sqrt{55}}{2}
$