use the quadratic equation: (-b±√(b^2-4ac))/2a where a=(k-1), b=2, c=-(k-3)
Factor out a 4 and multiply out under the square root sign:
Now, square root of 4 is 2, and putting together like terms you have:
Now looking at just what is under the square root sign: k^2-4k+4, which will factor to (k-2)^2.
square root of a square, so you have (2±2(k-2))/2(k-1)
Factor out a 2: (1±(k-2))/(k-1)
So, since we were able to factor out of the square root sign without getting a negative number, there can be only real roots.
BUT at the end you have a (k-1) in the denominator. so if k=1, we do not have an answer. BUT now look at the original equation. When k=1, the x^2 term becomes 0 and we are left with 2x-(1-3)=0, or 2x+2=0 or x=-1. So even when k=1, we still have a real root.
Therefore for all values of k, we will have real roots.