1. **Problem:** Find the number of intersection points between the parabola $y=3x^2$ and the line $y=3x+5$. 2. **Set the equations equal to find intersection points:** $$3x^2 = 3x + 5$$ 3. **Rewrite the equation:** $$3x^2 - 3x - 5 = 0$$ 4. **Use the quadratic formula:** For $ax^2 + bx + c = 0$, solutions are $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $a=3$, $b=-3$, $c=-5$. 5. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = (-3)^2 - 4 \times 3 \times (-5) = 9 + 60 = 69$$ 6. **Since $\Delta > 0$, there are two distinct real roots, so the parabola and line intersect at 2 points.** **Final answer:** 2