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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4y^2+5y-6=0

a = 4; b = 5; c = -6;
Δ = b2-4ac
Δ = 52-4·4·(-6)
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{-(5)-11}{2*4}=\frac{-16}{8}
=-2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+11}{2*4}=\frac{6}{8}
=3/4
$