(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^2-1m-20=0

a = 1; b = -1; c = -20;
Δ = b2-4ac
Δ = -12-4·1·(-20)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-9}{2*1}=\frac{-8}{2}
=-4
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+9}{2*1}=\frac{10}{2}
=5
$