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

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

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

We multiply parentheses ..

(+y^2+4y+5y+20)=0

We get rid of parentheses

y^2+4y+5y+20=0

We add all the numbers together, and all the variables

y^2+9y+20=0

a = 1; b = 9; c = +20;
Δ = b2-4ac
Δ = 92-4·1·20
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-1}{2*1}=\frac{-10}{2}
=-5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+1}{2*1}=\frac{-8}{2}
=-4
$