(6y+4)(3y-5)=0

Solution for (6y+4)(3y-5)=0 equation:

(6y+4)(3y-5)=0

We multiply parentheses ..

(+18y^2-30y+12y-20)=0

We get rid of parentheses

18y^2-30y+12y-20=0

We add all the numbers together, and all the variables

18y^2-18y-20=0

a = 18; b = -18; c = -20;
Δ = b2-4ac
Δ = -182-4·18·(-20)
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-42}{2*18}=\frac{-24}{36}
=-2/3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+42}{2*18}=\frac{60}{36}
=1+2/3
$