Find the product of all positive integer values of $c$ such that $8x^2+15x+c=0$ has two real roots.
I know that a quadratic equation has two real roots if and only if its discriminant is positive. For this equation, the discriminant is $15^2-4(8)(c)$. So I need to solve the inequality $15^2-4(8)(c)>0$. Simplifying, I get $225-32c>0$. Dividing by 32, I get $7.03125-c>0$. Adding $c$ to both sides, I get $7.03125>c$. This means that $c$ has to be less than 7.03125, but also a positive integer. The only positive integers that satisfy this are 1, 2, 3, 4, 5, 6, and 7. To find the product of these values, I can multiply them all together. I get $1\times 2\times 3\times 4\times 5\times 6\times 7=5040$. 