2m(m-3)-(m-4)(m-5)=8

Solution for 2m(m-3)-(m-4)(m-5)=8 equation:

2m(m-3)-(m-4)(m-5)=8

We move all terms to the left:

2m(m-3)-(m-4)(m-5)-(8)=0

We multiply parentheses

2m^2-6m-(m-4)(m-5)-8=0

We multiply parentheses ..

2m^2-(+m^2-5m-4m+20)-6m-8=0

We get rid of parentheses

2m^2-m^2+5m+4m-6m-20-8=0

We add all the numbers together, and all the variables

m^2+3m-28=0

a = 1; b = 3; c = -28;
Δ = b2-4ac
Δ = 32-4·1·(-28)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-11}{2*1}=\frac{-14}{2}
=-7
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+11}{2*1}=\frac{8}{2}
=4
$