(8+y)(4y+3)=0

Solution for (8+y)(4y+3)=0 equation:

(8+y)(4y+3)=0

We add all the numbers together, and all the variables

(y+8)(4y+3)=0

We multiply parentheses ..

(+4y^2+3y+32y+24)=0

We get rid of parentheses

4y^2+3y+32y+24=0

We add all the numbers together, and all the variables

4y^2+35y+24=0

a = 4; b = 35; c = +24;
Δ = b2-4ac
Δ = 352-4·4·24
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-29}{2*4}=\frac{-64}{8}
=-8
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+29}{2*4}=\frac{-6}{8}
=-3/4
$