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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

3y^2+8y+6y+16=0

We add all the numbers together, and all the variables

3y^2+14y+16=0

a = 3; b = 14; c = +16;
Δ = b2-4ac
Δ = 142-4·3·16
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2}{2*3}=\frac{-16}{6}
=-2+2/3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2}{2*3}=\frac{-12}{6}
=-2
$