(4y-1)(5y-4)=0

Solution for (4y-1)(5y-4)=0 equation:

(4y-1)(5y-4)=0

We multiply parentheses ..

(+20y^2-16y-5y+4)=0

We get rid of parentheses

20y^2-16y-5y+4=0

We add all the numbers together, and all the variables

20y^2-21y+4=0

a = 20; b = -21; c = +4;
Δ = b2-4ac
Δ = -212-4·20·4
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-11}{2*20}=\frac{10}{40}
=1/4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+11}{2*20}=\frac{32}{40}
=4/5
$