3x^2+2x-k=0; (this was taken from the text, rather than the title).
Rewrite: x^2+2x/3=k/3. Halve the x term, then square the coefficient=1/9; now, add 1/9 to each side: x^2+2x/3+1/9=k/3+1/9.
Rewrite again: (x+1/3)^2=k/3+1/9. The right-hand side must be positive for real roots, so k/3+1/9>0, and k>-1/3.
If we take the title: 3x^2=k-2, so k-2>0, so k>2.