How does one find values of $m$ for which the roots of $2x^2-mx-8=0$ differ by $m-1$ 
Find values of $m$ for which the roots of $2x^2-mx-8$ differ by $m-1$.

When I attempted to solve this I tried to simplify it into something like this: $((m-1)+2)((m-1)-4)$ but when I expanded I got $m^2 -4m -5$ which is nowhere near $2x^2-mx-8$ so my question is how to do solve these types of problems. I thank you in advance for your help. :)
 A: The roots, by quadratic formula, are:
$$x=\frac{m\pm \sqrt{m^2+64}}{4}$$
The difference is
$$\frac{m+\sqrt{m^2+64}-(m-\sqrt{m^2+64})}{4}=\frac{\sqrt{m^2+64}}{2}$$
So $$2m-2=\sqrt{m^2+64}\implies 4m^2-8m+4=m^2+64\implies (3m+10)(m-6)=0$$
So $m=\frac{-10}{3}, 6$.
A: Another approach is to use the property of sum and product of roots (Vieta's Formula).
Say $\alpha$ and $\beta$ are roots to $2x^2-mx-8=0$, we have
$$\alpha+\beta=m/2\\
\alpha\beta=-4\\
\alpha-\beta=m-1$$
Eliminating $\alpha$ and $\beta$ gives $m=6,-10/3$.
A: We may also directly construct the polynomials we seek .  The parabola corresponding to $ \ y \ = \ 2x^2 - mx - 8 \ = \ 2·\left(x - \frac{m}{4} \right)^2 \ - \ \left(8 + \frac{m^2}{8} \right) \ $ has its vertex at $ \ x = \frac{m}{4} \ \ , $ which is the midpoint of its $ \ x-$ intercepts.  (These always exist, since this is an "upward-opening" parabola and the $ \ y-$coordinate of the vertex  is always negative .)
Since it is desired that the difference between the roots of $ \ y \ = \ 2x^2 - mx - 8 \ = \ 0 \ \ $ be $ \ (m - 1) \ \ , $ the roots are $ \ \frac{m}{4} \ \pm \ \frac{(m - 1)}{2} \ = \ \frac{3m}{4} - \frac{1}{2} \  ,  \ -\frac{m}{4} + \frac{1}{2} \ \ . $  A quadratic polynomial with these as zeroes is then
$$ 2·\left(x \ - \ \frac{3m}{4} \ + \ \frac{1}{2} \right)·\left(x \ + \ \frac{m}{4} \ - \ \frac{1}{2} \right) \ \ = \ \ 2x^2 \ - \ mx \ + \ \left(m \ - \ \frac{3m^2}{8} \ - \ \frac12 \right) \ \ . $$
As  $ \ -8 \ \  $ is the given constant term, we therefore have
$$ m \ - \ \frac{3m^2}{8} \ - \ \frac12 \ \ = \ \ -8 \ \ \Rightarrow \ \ 3m^2 \ - \ 8m \ - \ 60 \ \ = \ \ (3m \ + \ 10) \ · \ (m \ - \ 6) \ \ = \ \ 0 \ \ . $$
The two quadratic polynomials with the specified property are then
• $ \ \ \mathbf{m \ = \ 6 \ \ :} \quad 2x^2 \ - \ 6x \ - \ 8 \ \ $ with zeroes $ \ 4 \ \ , \ \ -1 \ \ \ $ [difference: $ \ \ 4 \ - \ (-1) \ = \ 6 \ - \ 1 $ ]
and
• $ \ \ \mathbf{m \ = \ -\frac{10}{3} \ \ :} \quad 2x^2 \ + \ \frac{10}{3}x \ - \ 8 \ \ $ with zeroes $ \ \frac43 \ \ , \ \ -3 \ \ \ $
[difference: $ \ \ (-3) \ - \ \frac43   \ = \ -\frac{13}{3} \ = \  -\frac{10}{3} \ - \ 1 $ ]   .
