(m-3)(m-4)=0

Solution for (m-3)(m-4)=0 equation:

(m-3)(m-4)=0

We multiply parentheses ..

(+m^2-4m-3m+12)=0

We get rid of parentheses

m^2-4m-3m+12=0

We add all the numbers together, and all the variables

m^2-7m+12=0

a = 1; b = -7; c = +12;
Δ = b2-4ac
Δ = -72-4·1·12
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{-(-7)-1}{2*1}=\frac{6}{2}
=3
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+1}{2*1}=\frac{8}{2}
=4
$