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

Solution for (m-2)(m-5)=0 equation:

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

m^2-7m+10=0

a = 1; b = -7; c = +10;
Δ = b2-4ac
Δ = -72-4·1·10
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-3}{2*1}=\frac{4}{2}
=2
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+3}{2*1}=\frac{10}{2}
=5
$