(4y-3)(5y-1)=0

Solution for (4y-3)(5y-1)=0 equation:

(4y-3)(5y-1)=0

We multiply parentheses ..

(+20y^2-4y-15y+3)=0

We get rid of parentheses

20y^2-4y-15y+3=0

We add all the numbers together, and all the variables

20y^2-19y+3=0

a = 20; b = -19; c = +3;
Δ = b2-4ac
Δ = -192-4·20·3
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{-(-19)-11}{2*20}=\frac{8}{40}
=1/5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+11}{2*20}=\frac{30}{40}
=3/4
$