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

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

(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
$