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

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

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

We multiply parentheses ..

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

We get rid of parentheses

4y^2-32y+5y-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{-10}{8}
=-1+1/4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+37}{2*4}=\frac{64}{8}
=8
$