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

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

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

We multiply parentheses

4y^2+5y=0

a = 4; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·4·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*4}=\frac{-10}{8}
=-1+1/4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*4}=\frac{0}{8}
=0
$