Induction principle: $n^2-7n+12≥0$ for every $n≥3$ How can I prove that $n^2-7n+12≥0$ for every $n≥3$?

I know that for $n=3$ I have $0≥0$ so the inductive Hypothesis is true.

Now for $n+1$ I have $(n+1)^2-7(n+1)+12=n^2-5n+6$ and now I don't know how to go on...
HINT

$(n+1)^2-7(n+1)+12=n^2-5n+6= (n^2 -7n+12)+(2n-6)$
Use completing the square:

$$n^2-7n+12=\bigg(n-\frac{7}{2}\bigg)^2-\bigg(\frac{7}{2}\bigg)^2+12$$

$$\to\bigg(n-\frac{7}{2}\bigg)^2-\frac{49}{4}+\frac{48}{4}=\bigg(n-\frac{7}{2}\bigg)^2-\frac{1}{4}$$

Note that $k^2 \ge 0$ $\forall k \in \Bbb R$, and for $n\ge 3, \bigg(n-\frac{7}{2}\bigg)^2\ge\frac{1}{4}$, hence $\bigg(n-\frac{7}{2}\bigg)^2-\frac{1}{4}\ge 0$