(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+11y-3=0

a = 20; b = 11; c = -3;
Δ = b2-4ac
Δ = 112-4·20·(-3)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-19}{2*20}=\frac{-30}{40}
=-3/4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+19}{2*20}=\frac{8}{40}
=1/5
$