Find the product of all positive integer values of $c$ such that the quadratic equation $3x^2+7x+c=15x-10$ has two real roots.

Answer 1

We want the quadratic formula to yield two real roots, so the discriminant must be greater than $0.$ This gives us
\[b^2 - 4ac = 7^2 - 4 \cdot 3(c - 15) > 0.\]Then $49 - 12c + 180 > 0,$ so $c < \frac{229}{12}.$

Also, $c = 15x - 10 - 3x^2 - 7x = -3x^2 + 8x - 10,$ so the quadratic $-3x^2 + 8x + (10 - c) = 0$ must have two real roots.

A quadratic has two real roots if and only if its discriminant is nonnegative, so
\[8^2 - 4(-3)(10 - c) \ge 0.\]Then $64 - 12(10 - c) \ge 0,$ so $c \ge 8.$

The possible values of $c$ are then 8, 9, 10, $\dots,$ 18. The product of all positive integer values of $c$ from 8 to $\left\lfloor \frac{229}{12} \right\rfloor = 19$ is $\boxed{901,536}.$