2(w+3)w=228

Solution for 2(w+3)w=228 equation:

2(w+3)w=228

We move all terms to the left:

2(w+3)w-(228)=0

We multiply parentheses

2w^2+6w-228=0

a = 2; b = 6; c = -228;
Δ = b2-4ac
Δ = 62-4·2·(-228)
Δ = 1860
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:

$\sqrt{\Delta}=\sqrt{1860}=\sqrt{4*465}=\sqrt{4}*\sqrt{465}=2\sqrt{465}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{465}}{2*2}=\frac{-6-2\sqrt{465}}{4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{465}}{2*2}=\frac{-6+2\sqrt{465}}{4}
$