20b^2=4b

Solution for 20b^2=4b equation:

20b^2=4b

We move all terms to the left:

20b^2-(4b)=0

a = 20; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·20·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*20}=\frac{0}{40}
=0
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*20}=\frac{8}{40}
=1/5
$