quadratic

Simplify to get:

$x^2-mx+20=0$

Now, for this Quadratic to have no real solutions, it should not intersect the x-axis. 

So, the discriminant , which is: $b^2-4ac$ must be less than zero.

So:
$b^2-4ac<0$

$\iff m^2-80 <0$  (Now we have to solve this Quadratic inequality for m).

The critical values:  $m=\pm 4\sqrt{5}$

Next, sketch (*) and place the critical points, you want the region below the x-axis (Hence, between the critical values):
Thus, \(-4\sqrt{5}

Since hte question is asking about the largest integer m, let us convert this to decimal:

\(-8.94427..

Therefore, the largest integer m, is: $m=8$ is the desired solution.

Hope this helps!