3w+63=2w^2-18w

Solution for 3w+63=2w^2-18w equation:

3w+63=2w^2-18w

We move all terms to the left:

3w+63-(2w^2-18w)=0

We get rid of parentheses

-2w^2+3w+18w+63=0

We add all the numbers together, and all the variables

-2w^2+21w+63=0

a = -2; b = 21; c = +63;
Δ = b2-4ac
Δ = 212-4·(-2)·63
Δ = 945
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:

$\sqrt{\Delta}=\sqrt{945}=\sqrt{9*105}=\sqrt{9}*\sqrt{105}=3\sqrt{105}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-3\sqrt{105}}{2*-2}=\frac{-21-3\sqrt{105}}{-4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+3\sqrt{105}}{2*-2}=\frac{-21+3\sqrt{105}}{-4}
$