m(m-3)=2

Solution for m(m-3)=2 equation:

m(m-3)=2

We move all terms to the left:

m(m-3)-(2)=0

We multiply parentheses

m^2-3m-2=0

a = 1; b = -3; c = -2;
Δ = b2-4ac
Δ = -32-4·1·(-2)
Δ = 17
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{17}}{2*1}=\frac{3-\sqrt{17}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{17}}{2*1}=\frac{3+\sqrt{17}}{2}
$