3w^2+32w=424

Solution for 3w^2+32w=424 equation:

3w^2+32w=424

We move all terms to the left:

3w^2+32w-(424)=0

a = 3; b = 32; c = -424;
Δ = b2-4ac
Δ = 322-4·3·(-424)
Δ = 6112
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:

$\sqrt{\Delta}=\sqrt{6112}=\sqrt{16*382}=\sqrt{16}*\sqrt{382}=4\sqrt{382}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-4\sqrt{382}}{2*3}=\frac{-32-4\sqrt{382}}{6}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+4\sqrt{382}}{2*3}=\frac{-32+4\sqrt{382}}{6}
$