A=-3w^2+480w

Solution for A=-3w^2+480w equation:

=-3A^2+480A

We move all terms to the left:

-(-3A^2+480A)=0

We get rid of parentheses

3A^2-480A=0

a = 3; b = -480; c = 0;
Δ = b2-4ac
Δ = -4802-4·3·0
Δ = 230400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{230400}=480$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-480)-480}{2*3}=\frac{0}{6}
=0
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-480)+480}{2*3}=\frac{960}{6}
=160
$