80=40w-w^2

Solution for 80=40w-w^2 equation:

80=40w-w^2

We move all terms to the left:

80-(40w-w^2)=0

We get rid of parentheses

w^2-40w+80=0

a = 1; b = -40; c = +80;
Δ = b2-4ac
Δ = -402-4·1·80
Δ = 1280
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:

$\sqrt{\Delta}=\sqrt{1280}=\sqrt{256*5}=\sqrt{256}*\sqrt{5}=16\sqrt{5}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-16\sqrt{5}}{2*1}=\frac{40-16\sqrt{5}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+16\sqrt{5}}{2*1}=\frac{40+16\sqrt{5}}{2}
$