(3-4y)(2y+1)=0

Solution for (3-4y)(2y+1)=0 equation:

(3-4y)(2y+1)=0

We add all the numbers together, and all the variables

(-4y+3)(2y+1)=0

We multiply parentheses ..

(-8y^2-4y+6y+3)=0

We get rid of parentheses

-8y^2-4y+6y+3=0

We add all the numbers together, and all the variables

-8y^2+2y+3=0

a = -8; b = 2; c = +3;
Δ = b2-4ac
Δ = 22-4·(-8)·3
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-10}{2*-8}=\frac{-12}{-16}
=3/4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+10}{2*-8}=\frac{8}{-16}
=-1/2
$