(2y+9)(3y-2)=0

Solution for (2y+9)(3y-2)=0 equation:

(2y+9)(3y-2)=0

We multiply parentheses ..

(+6y^2-4y+27y-18)=0

We get rid of parentheses

6y^2-4y+27y-18=0

We add all the numbers together, and all the variables

6y^2+23y-18=0

a = 6; b = 23; c = -18;
Δ = b2-4ac
Δ = 232-4·6·(-18)
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-31}{2*6}=\frac{-54}{12}
=-4+1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+31}{2*6}=\frac{8}{12}
=2/3
$