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

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

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

We add all the numbers together, and all the variables

(3y+4)(-1y+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{-12}{-6}
=+2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+10}{2*-3}=\frac{8}{-6}
=-1+1/3
$