324=w^2+3w

Solution for 324=w^2+3w equation:

324=w^2+3w

We move all terms to the left:

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

We get rid of parentheses

-w^2-3w+324=0

We add all the numbers together, and all the variables

-1w^2-3w+324=0

a = -1; b = -3; c = +324;
Δ = b2-4ac
Δ = -32-4·(-1)·324
Δ = 1305
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:

$\sqrt{\Delta}=\sqrt{1305}=\sqrt{9*145}=\sqrt{9}*\sqrt{145}=3\sqrt{145}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3\sqrt{145}}{2*-1}=\frac{3-3\sqrt{145}}{-2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3\sqrt{145}}{2*-1}=\frac{3+3\sqrt{145}}{-2}
$