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

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

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

We add all the numbers together, and all the variables

(3y+4)(-1y+6)=0

We multiply parentheses ..

(-3y^2+18y-4y+24)=0

We get rid of parentheses

-3y^2+18y-4y+24=0

We add all the numbers together, and all the variables

-3y^2+14y+24=0

a = -3; b = 14; c = +24;
Δ = b2-4ac
Δ = 142-4·(-3)·24
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-22}{2*-3}=\frac{-36}{-6}
=+6
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+22}{2*-3}=\frac{8}{-6}
=-1+1/3
$