the quadratic equation $3(k+2)x^2+(k+5)x+k=0$ has real roots
show $(k-1)(11k+25) \geq 0 $
If $\Delta$ greater than $0$ it has real roots so,
$$\Delta = (k+5)^2 - 4 \cdot (3(k+2))\cdot k$$
$$k^2+10k+25-4(3k+6)\cdot k = (?)-12k^2-24k$$
which doesn't help and the answer is not the same for Wolfram|Alpha.
So how can I prove this is greater than $0$?