1. **State the problem:** Solve the quadratic equation $$c t^2 + 15t = -36$$ for $t$. 2. **Rewrite the equation:** Move all terms to one side to set the equation to zero: $$c t^2 + 15t + 36 = 0$$ 3. **Use the quadratic formula:** For an equation $$a t^2 + b t + c = 0$$, the solutions are given by $$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $a = c$, $b = 15$, and $c = 36$ (constant term). 4. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 15^2 - 4 \times c \times 36 = 225 - 144c$$ 5. **Write the solutions:** $$t = \frac{-15 \pm \sqrt{225 - 144c}}{2c}$$ 6. **Interpretation:** The solutions depend on the value of $c$. For real solutions, the discriminant must be non-negative: $$225 - 144c \geq 0 \implies c \leq \frac{225}{144} = \frac{25}{16} = 1.5625$$ **Final answer:** $$t = \frac{-15 \pm \sqrt{225 - 144c}}{2c}$$