20b-b^2=40

Solution for 20b-b^2=40 equation:

20b-b^2=40

We move all terms to the left:

20b-b^2-(40)=0

We add all the numbers together, and all the variables

-1b^2+20b-40=0

a = -1; b = 20; c = -40;
Δ = b2-4ac
Δ = 202-4·(-1)·(-40)
Δ = 240
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{15}}{2*-1}=\frac{-20-4\sqrt{15}}{-2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{15}}{2*-1}=\frac{-20+4\sqrt{15}}{-2}
$