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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

3y^2+24y-2y-16=0

We add all the numbers together, and all the variables

3y^2+22y-16=0

a = 3; b = 22; c = -16;
Δ = b2-4ac
Δ = 222-4·3·(-16)
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-26}{2*3}=\frac{-48}{6}
=-8
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+26}{2*3}=\frac{4}{6}
=2/3
$