w^2+3w=540

Solution for w^2+3w=540 equation:

w^2+3w=540

We move all terms to the left:

w^2+3w-(540)=0

a = 1; b = 3; c = -540;
Δ = b2-4ac
Δ = 32-4·1·(-540)
Δ = 2169
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:

$\sqrt{\Delta}=\sqrt{2169}=\sqrt{9*241}=\sqrt{9}*\sqrt{241}=3\sqrt{241}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{241}}{2*1}=\frac{-3-3\sqrt{241}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{241}}{2*1}=\frac{-3+3\sqrt{241}}{2}
$