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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

6y^2-5y-4=0

a = 6; b = -5; c = -4;
Δ = b2-4ac
Δ = -52-4·6·(-4)
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*6}=\frac{-6}{12}
=-1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+11}{2*6}=\frac{16}{12}
=1+1/3
$