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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

3y^2-2y-8=0

a = 3; b = -2; c = -8;
Δ = b2-4ac
Δ = -22-4·3·(-8)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-10}{2*3}=\frac{-8}{6}
=-1+1/3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+10}{2*3}=\frac{12}{6}
=2
$