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

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

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

We multiply parentheses ..

(+6y^2-9y+8y-12)=0

We get rid of parentheses

6y^2-9y+8y-12=0

We add all the numbers together, and all the variables

6y^2-1y-12=0

a = 6; b = -1; c = -12;
Δ = b2-4ac
Δ = -12-4·6·(-12)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-17}{2*6}=\frac{-16}{12}
=-1+1/3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+17}{2*6}=\frac{18}{12}
=1+1/2
$