(8+y)(4y-5)=0

Solution for (8+y)(4y-5)=0 equation:

(8+y)(4y-5)=0

We add all the numbers together, and all the variables

(y+8)(4y-5)=0

We multiply parentheses ..

(+4y^2-5y+32y-40)=0

We get rid of parentheses

4y^2-5y+32y-40=0

We add all the numbers together, and all the variables

4y^2+27y-40=0

a = 4; b = 27; c = -40;
Δ = b2-4ac
Δ = 272-4·4·(-40)
Δ = 1369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1369}=37$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-37}{2*4}=\frac{-64}{8}
=-8
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+37}{2*4}=\frac{10}{8}
=1+1/4
$