(m-4)(m-5)=8

Solution for (m-4)(m-5)=8 equation:

(m-4)(m-5)=8

We move all terms to the left:

(m-4)(m-5)-(8)=0

We multiply parentheses ..

(+m^2-5m-4m+20)-8=0

We get rid of parentheses

m^2-5m-4m+20-8=0

We add all the numbers together, and all the variables

m^2-9m+12=0

a = 1; b = -9; c = +12;
Δ = b2-4ac
Δ = -92-4·1·12
Δ = 33
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-\sqrt{33}}{2*1}=\frac{9-\sqrt{33}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{33}}{2*1}=\frac{9+\sqrt{33}}{2}
$