Find the product of all positive integer values of $c$ such that $3x^2+7x+c=0$ has two real roots.
To find the product of all positive integer values of $c$, I need to first find all such values of $c$. I know that a quadratic equation of the form $ax^2+bx+c=0$ has two real roots if and only if its discriminant, $b^2-4ac$, is positive. So, in this case, I need to find all positive integers $c$ such that $7^2-4\cdot 3\cdot c>0$. Simplifying, I get $49-12c>0$, or $c<\frac{49}{12}$. Since $c$ has to be a positive integer, the only possible values of $c$ are $1, 2, 3$, and $4$. To find the product of these values, I just need to multiply them together. The product is $1\cdot 2\cdot 3\cdot 4=24$.

