3(y-4)(y+2)+(2y-1)(y+8)=0

Solution for 3(y-4)(y+2)+(2y-1)(y+8)=0 equation:

3(y-4)(y+2)+(2y-1)(y+8)=0

We multiply parentheses ..

3(+y^2+2y-4y-8)+(2y-1)(y+8)=0

We multiply parentheses

3y^2+6y-12y+(2y-1)(y+8)-24=0

We multiply parentheses ..

3y^2+(+2y^2+16y-1y-8)+6y-12y-24=0

We add all the numbers together, and all the variables

3y^2+(+2y^2+16y-1y-8)-6y-24=0

We get rid of parentheses

3y^2+2y^2+16y-1y-6y-8-24=0

We add all the numbers together, and all the variables

5y^2+9y-32=0

a = 5; b = 9; c = -32;
Δ = b2-4ac
Δ = 92-4·5·(-32)
Δ = 721
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-\sqrt{721}}{2*5}=\frac{-9-\sqrt{721}}{10}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{721}}{2*5}=\frac{-9+\sqrt{721}}{10}
$