Your solution is somewhat correct, just a bit complex. 

For no real roots, only need to prove that D < 0,  of the resulting quadratic equation (same as all the other resulting quadratic equations in the earlier questions)

x ² + x + (p² /2) = px – 3

x ² + (1-p)x +  (p² /2) +3 = 0

a = 1, b = (1-p) ,  c =  (p² /2) +3

D = b² – 4ac = (1-p)²  – 4(1)[(p² /2) +3]

                       = 1 -p² -2p -2p² -12

                        = – p²  -2p -11

                        = – [ p²  + 2p + 11 ]       

                        = – [ (p²  + 2p + 1) – 1 + 11 ]  (complete the square)

                        = – [(p+1)²  + 10 ]

Since  [(p+1)² + 10 ] is a positive number for all values of p,  having a negative sign makes  – [(p+1)²  + 10 ] a negative number.

Hence  D = – [(p+1)²  + 10 ] < 0 implies that x ² + x + (p² /2) = px – 3 has no real roots for all values of p.

Got it, thanks.