4(y+1)(4y-5)=0

Solution for 4(y+1)(4y-5)=0 equation:

4(y+1)(4y-5)=0

We multiply parentheses ..

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

We multiply parentheses

16y^2-20y+16y-20=0

We add all the numbers together, and all the variables

16y^2-4y-20=0

a = 16; b = -4; c = -20;
Δ = b2-4ac
Δ = -42-4·16·(-20)
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-36}{2*16}=\frac{-32}{32}
=-1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+36}{2*16}=\frac{40}{32}
=1+1/4
$