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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-3y^2-5y+8=0

a = -3; b = -5; c = +8;
Δ = b2-4ac
Δ = -52-4·(-3)·8
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-11}{2*-3}=\frac{-6}{-6}
=1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+11}{2*-3}=\frac{16}{-6}
=-2+2/3
$