1. **State the problem:** We are given the function $f(x) = 3x^2 + 4x - k + 3$ and the condition $f(-1) = -2$. We need to find: (a) The value of $k$. (b) The value(s) of $t$ such that $f(t) = 38$. 2. **Find $k$ using $f(-1) = -2$:** Substitute $x = -1$ into the function: $$f(-1) = 3(-1)^2 + 4(-1) - k + 3 = 3(1) - 4 - k + 3 = 3 - 4 - k + 3$$ Simplify: $$3 - 4 - k + 3 = (3 - 4 + 3) - k = 2 - k$$ Given $f(-1) = -2$, set equal: $$2 - k = -2$$ Solve for $k$: $$2 - k = -2$$ $$\Rightarrow \cancel{2} - k = \cancel{-2}$$ $$-k = -4$$ Multiply both sides by $-1$: $$k = 4$$ 3. **Rewrite the function with $k=4$:** $$f(x) = 3x^2 + 4x - 4 + 3 = 3x^2 + 4x - 1$$ 4. **Find $t$ such that $f(t) = 38$:** Set the function equal to 38: $$3t^2 + 4t - 1 = 38$$ Bring all terms to one side: $$3t^2 + 4t - 1 - 38 = 0$$ $$3t^2 + 4t - 39 = 0$$ 5. **Solve the quadratic equation:** Use the quadratic formula: $$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=3$, $b=4$, $c=-39$. Calculate the discriminant: $$\Delta = b^2 - 4ac = 4^2 - 4(3)(-39) = 16 + 468 = 484$$ Calculate the square root: $$\sqrt{484} = 22$$ Calculate the roots: $$t = \frac{-4 \pm 22}{2 \times 3} = \frac{-4 \pm 22}{6}$$ Two solutions: $$t_1 = \frac{-4 + 22}{6} = \frac{18}{6} = 3$$ $$t_2 = \frac{-4 - 22}{6} = \frac{-26}{6} = -\frac{13}{3}$$ **Final answers:** (a) $k = 4$ (b) $t = 3$ or $t = -\frac{13}{3}$