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

Solution for (4m-5)(m-5)=0 equation:

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

We multiply parentheses ..

(+4m^2-20m-5m+25)=0

We get rid of parentheses

4m^2-20m-5m+25=0

We add all the numbers together, and all the variables

4m^2-25m+25=0

a = 4; b = -25; c = +25;
Δ = b2-4ac
Δ = -252-4·4·25
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-15}{2*4}=\frac{10}{8}
=1+1/4
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+15}{2*4}=\frac{40}{8}
=5
$