(m+4)(m+5)=0

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

(m+4)(m+5)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

m^2+9m+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:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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