(m-2)(m-5)=4

Solution for (m-2)(m-5)=4 equation:

(m-2)(m-5)=4

We move all terms to the left:

(m-2)(m-5)-(4)=0

We multiply parentheses ..

(+m^2-5m-2m+10)-4=0

We get rid of parentheses

m^2-5m-2m+10-4=0

We add all the numbers together, and all the variables

m^2-7m+6=0

a = 1; b = -7; c = +6;
Δ = b2-4ac
Δ = -72-4·1·6
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-5}{2*1}=\frac{2}{2}
=1
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+5}{2*1}=\frac{12}{2}
=6
$