0=(m-1)(2m-5)=0

Solution for 0=(m-1)(2m-5)=0 equation:

0=(m-1)(2m-5)=0

We move all terms to the left:

0-((m-1)(2m-5))=0

We add all the numbers together, and all the variables

-((m-1)(2m-5))=0

We multiply parentheses ..

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


We calculate terms in parentheses: -((+2m^2-5m-2m+5)), so:

(+2m^2-5m-2m+5)

We get rid of parentheses

2m^2-5m-2m+5

We add all the numbers together, and all the variables

2m^2-7m+5

Back to the equation:

-(2m^2-7m+5)


We get rid of parentheses

-2m^2+7m-5=0

a = -2; b = 7; c = -5;
Δ = b2-4ac
Δ = 72-4·(-2)·(-5)
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*-2}=\frac{-10}{-4}
=2+1/2
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+3}{2*-2}=\frac{-4}{-4}
=1
$