Your solution is too complicated.

3-k = x² + (1-2k) x

x² + (1-2k) x – (3-k) =0

To find that this equation has 2 real and distinct roots, we need to prove that D > 0.

a = 1, b = (1-2k) , c = -(3-k) = k-3

(1-2k)² – 4(1)(k-3) = 1 – 4k + 4k² – 12 -4k

                                  = 4k² -8k + 13

To complete the square since it cannot be factorised,

                                  = k²  – 2k + 13/4  

                                  = k²  – 2k  + 13/4

                                  = k² -2k + 1  -1 +13/4  

                                  = (k-1)² + 9/4

So (k-1)² and 9/4 are both positive numbers for all values of k, it implies that (k-1)² + 9/4 > 0, hence D > 0.

Therefore,  3-k = x² + (1-2k) x  has two real and distinct roots for all values of k.

Understand, thanks.