Find the range of values of $k$ for which $3x^2-4(k-x)+2$ is always positive for all real values of $x$.
I've tried simplifying, until I got to: $3x^2-4k+4x+2$.
Since it must always be positive, the discriminant, $b^2-4ac$ must be negative, ie $b^2-4ac<0$.
$b^2-4ac<0$
$(4)^2-4(3)(2-4k)<0$
$16-24+48k<0$
$48k<8$
$k<\frac 16$
I got stuck here. Is anyone able to help me continue?
EDIT: Realised that there was a typo, the discriminant should be negative, thank you Deepak.
You seem to have already got the answer (which is simply $k < \frac 16$). There is only one additional modification I would make to your answer. That is to assert that the coefficient of $x^2$ is positive (in this case $3$). If it were negative, you would never be able to satisfy the required condition since for high enough values of $x$, the quadratic expression would always be negative.
EDIT: just noticed you wrote that the discriminant must be positive. I hope that's just a typo, and not a conceptual error. It seems to have been just a typo because you set up the inequality correctly. Anyway, remember that you need complex roots for a quadratic curve to lie wholly on one side of the $x$-axis or the other (which side depends on the sign of the leading coefficient). Complex roots means the discriminant is negative).