1. **State the problem:** We have a function $f(x) = (x + 3)^2 + 5$ and need to find $f(0)$ and rewrite $f(x)$ in standard quadratic form. Then solve the quadratic equation $3x^2 - 4x - 9 = 0$. 2. **Find $f(0)$:** Substitute $x=0$ into $f(x)$: $$f(0) = (0 + 3)^2 + 5 = 3^2 + 5 = 9 + 5 = 14$$ 3. **Rewrite $f(x)$ in standard form $ax^2 + bx + c$:** Expand the square: $$f(x) = (x + 3)^2 + 5 = (x^2 + 2 \times 3 \times x + 3^2) + 5 = x^2 + 6x + 9 + 5 = x^2 + 6x + 14$$ So, $a=1$, $b=6$, $c=14$. 4. **Solve $3x^2 - 4x - 9 = 0$ using the quadratic formula:** The quadratic formula is: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $a=3$, $b=-4$, $c=-9$. Calculate the discriminant: $$\Delta = b^2 - 4ac = (-4)^2 - 4 \times 3 \times (-9) = 16 + 108 = 124$$ Calculate the roots: $$x = \frac{-(-4) \pm \sqrt{124}}{2 \times 3} = \frac{4 \pm \sqrt{124}}{6}$$ Simplify $\sqrt{124}$: $$\sqrt{124} = \sqrt{4 \times 31} = 2\sqrt{31}$$ So, $$x = \frac{4 \pm 2\sqrt{31}}{6} = \frac{\cancel{2} \times 2 \pm \cancel{2} \times \sqrt{31}}{\cancel{2} \times 3} = \frac{2 \pm \sqrt{31}}{3}$$ Calculate decimal approximations: $$\sqrt{31} \approx 5.57$$ Therefore, $$x_1 = \frac{2 + 5.57}{3} = \frac{7.57}{3} \approx 2.52$$ $$x_2 = \frac{2 - 5.57}{3} = \frac{-3.57}{3} \approx -1.19$$ **Final answers:** - $f(0) = 14$ - $f(x) = x^2 + 6x + 14$ - Solutions to $3x^2 - 4x - 9 = 0$ are $x \approx 2.52$ and $x \approx -1.19$