(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)(y+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+11y+8=0

a = 3; b = 11; c = +8;
Δ = b2-4ac
Δ = 112-4·3·8
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-5}{2*3}=\frac{-16}{6}
=-2+2/3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+5}{2*3}=\frac{-6}{6}
=-1
$