How do I determine the values of $k$ for which $3x^2 + kx+12 = 0$ has no real solutions, $1$ real solution, and $2$ real solutions? I know that a point has to be $(0,12)$ just because of the $C$ term always being the y-intercept. 
Please just tell me how to approach the problem/give tips, so that you are not doing my homework for me. 

By the distributive property the equation is equivalently:
$$0=3\left(x^2+\frac{k}{3}x\right)+12$$
Subtracting $12$ from both sides of the equation.
$$-12=3\left(x^2+\frac{k}{3}x\right)$$
Dividing by $3$ on both sides of the equation.
$$-4=x^2+\frac{k}{3}x$$
Using the fact that $\left(x+\frac{y}{2}\right)^2=x^2+2\left(x\right)\left(\frac{y}{2}\right)+\frac{y^2}{4}$:
$$-4=\left(x+\frac{k}{6}\right)^2-\frac{k^2}{36}$$
$$\frac{k^2}{36}-4=\left(x+\frac{k}{6}\right)^2$$
Note now that if $u^2=c$ then $c=0$ gives one solution $0$, $c>0$ gives $2$ solutions $\pm \sqrt{c}$, and $c<0$ gives no real solutions.
With a little more thinking we can see that If we want $1$ real root we need:
$$\frac{k^2}{36}-4=0$$
If we want $2$ real solutions we need:
$$\frac{k^2}{36}-4>0$$
And finally no real solutions:
$$\frac{k^2}{36}-4<0$$
