3600(w)=-w^2+120w

Solution for 3600(w)=-w^2+120w equation:

3600(w)=-w^2+120w

We move all terms to the left:

3600(w)-(-w^2+120w)=0

We get rid of parentheses

w^2-120w+3600w=0

We add all the numbers together, and all the variables

w^2+3480w=0

a = 1; b = 3480; c = 0;
Δ = b2-4ac
Δ = 34802-4·1·0
Δ = 12110400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12110400}=3480$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3480)-3480}{2*1}=\frac{-6960}{2}
=-3480
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3480)+3480}{2*1}=\frac{0}{2}
=0
$