(m-4)(m-10)=0

Solution for (m-4)(m-10)=0 equation:

(m-4)(m-10)=0

We multiply parentheses ..

(+m^2-10m-4m+40)=0

We get rid of parentheses

m^2-10m-4m+40=0

We add all the numbers together, and all the variables

m^2-14m+40=0

a = 1; b = -14; c = +40;
Δ = b2-4ac
Δ = -142-4·1·40
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-6}{2*1}=\frac{8}{2}
=4
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+6}{2*1}=\frac{20}{2}
=10
$