I understand how to solve this problem, but my method/logic seems a round-about way of doing so. Any tips on how to solve this question faster/more easily would be appreciated :)
Solving for part a) no solutions:
Find the discriminant: 16m^2 - 80
For no solutions: 16m^2 - 80 < 0
When 16m^2 -80 = 0, m = 5, -5. The interval between 5 and -5 is either below the x-axis (less than 0, i.e. no solutions) or above the x-axis (more than 0, i.e. 2 solutions).
Substituting a random number (0) between 5 and -5 into the discriminant gives -80 (less than 0) so we can confirm that -5 < m < 5 is the range of m required to give no solutions.
For part b) I apply the same logic, so 2 solutions obtained when m < -root5 and m > root 5
This is from the Cambridge Methods Unit 1 textbook.
First, you can use the reduced discriminant $\Delta'=4m^2-20=4(m^2-5)$. Next, there are no roots, as you said, if and only if $$\Delta'<0\iff m^2<5\iff |m|<\sqrt 5\iff -\sqrt 5<m<\sqrt 5.$$ Similarly for the existence of two roots: it amounts to solving $|m|>\sqrt 5$.