Show that the equation $2x^2+(6+p)x+3p=0$ has real roots for all real values of $p$

Question

Show that the equation $2x^2+(6+p)x+3p=0$ has real roots for all real values of $p$.

My attempt,

$b^2-4ac=(6+p)^2-4(2)(3p)$

$=p^2-12p+36 $

$(p-6)^2 \geq 0$

Can I prove like this? Thanks in advance.

Answer 1

You have one mistake in your solution.

In question term is $(6-p)$. But when you using it in formula you wrote it as $(6+p)$.

But your method to solve is correct. For two distinct real roots $b^2-4ac>0$