20y=16y^2

Solution for 20y=16y^2 equation:

20y=16y^2

We move all terms to the left:

20y-(16y^2)=0

determiningTheFunctionDomain
-16y^2+20y=0

a = -16; b = 20; c = 0;
Δ = b2-4ac
Δ = 202-4·(-16)·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20}{2*-16}=\frac{-40}{-32}
=1+1/4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20}{2*-16}=\frac{0}{-32}
=0
$