(8+y)(4y+2)=0

Solution for (8+y)(4y+2)=0 equation:

(8+y)(4y+2)=0

We add all the numbers together, and all the variables

(y+8)(4y+2)=0

We multiply parentheses ..

(+4y^2+2y+32y+16)=0

We get rid of parentheses

4y^2+2y+32y+16=0

We add all the numbers together, and all the variables

4y^2+34y+16=0

a = 4; b = 34; c = +16;
Δ = b2-4ac
Δ = 342-4·4·16
Δ = 900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{900}=30$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-30}{2*4}=\frac{-64}{8}
=-8
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+30}{2*4}=\frac{-4}{8}
=-1/2
$