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

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

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

We add all the numbers together, and all the variables

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