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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+3y^2+24y+y+8)=0

We get rid of parentheses

3y^2+24y+y+8=0

We add all the numbers together, and all the variables

3y^2+25y+8=0

a = 3; b = 25; c = +8;
Δ = b2-4ac
Δ = 252-4·3·8
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-23}{2*3}=\frac{-48}{6}
=-8
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+23}{2*3}=\frac{-2}{6}
=-1/3
$