1. The problem is to find the inverse Laplace transform of the function $$F(s) = \frac{10s + 23}{5s^2 + 7s + 12}$$ using the Laplace transform table. 2. Recall that the Laplace transform of a function $f(t)$ is defined as $$F(s) = \mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} f(t) dt$$ and the inverse Laplace transform is denoted as $$f(t) = \mathcal{L}^{-1}\{F(s)\}$$. 3. To find $f(t)$, we first simplify and factor the denominator if possible. The denominator is $$5s^2 + 7s + 12$$. 4. Calculate the discriminant: $$\Delta = 7^2 - 4 \times 5 \times 12 = 49 - 240 = -191$$ which is negative, so the quadratic has complex roots. 5. Complete the square or rewrite the denominator in a form suitable for inverse Laplace transform: $$5s^2 + 7s + 12 = 5\left(s^2 + \frac{7}{5}s + \frac{12}{5}\right)$$ 6. Complete the square inside the parentheses: $$s^2 + \frac{7}{5}s + \frac{12}{5} = \left(s + \frac{7}{10}\right)^2 + \frac{12}{5} - \left(\frac{7}{10}\right)^2 = \left(s + \frac{7}{10}\right)^2 + \frac{12}{5} - \frac{49}{100} = \left(s + \frac{7}{10}\right)^2 + \frac{240}{100} - \frac{49}{100} = \left(s + \frac{7}{10}\right)^2 + \frac{191}{100}$$ 7. So the denominator is: $$5\left(s + \frac{7}{10}\right)^2 + 5 \times \frac{191}{100} = 5\left(s + \frac{7}{10}\right)^2 + \frac{191}{20}$$ 8. Rewrite the numerator in terms of $s + \frac{7}{10}$: $$10s + 23 = 10\left(s + \frac{7}{10}\right) + 23 - 7 = 10\left(s + \frac{7}{10}\right) + 16$$ 9. Thus, $$F(s) = \frac{10\left(s + \frac{7}{10}\right) + 16}{5\left(s + \frac{7}{10}\right)^2 + \frac{191}{20}} = \frac{10\left(s + \frac{7}{10}\right)}{5\left(s + \frac{7}{10}\right)^2 + \frac{191}{20}} + \frac{16}{5\left(s + \frac{7}{10}\right)^2 + \frac{191}{20}}$$ 10. Let $a = \frac{7}{10}$ and $b = \sqrt{\frac{191}{100}} = \frac{\sqrt{191}}{10}$. 11. Using the Laplace transform table: - $$\mathcal{L}\{e^{-at} \cos(bt)\} = \frac{s + a}{(s + a)^2 + b^2}$$ - $$\mathcal{L}\{e^{-at} \sin(bt)\} = \frac{b}{(s + a)^2 + b^2}$$ 12. Rewrite $F(s)$ as: $$F(s) = 2 \times \frac{5\left(s + a\right)}{5\left(s + a\right)^2 + 5b^2} + \frac{16}{5\left(s + a\right)^2 + 5b^2}$$ 13. Factor out 5 in denominator: $$F(s) = 2 \times \frac{s + a}{(s + a)^2 + b^2} + \frac{16/5}{(s + a)^2 + b^2}$$ 14. Express the second term as: $$\frac{16/5}{(s + a)^2 + b^2} = \frac{16}{5b} \times \frac{b}{(s + a)^2 + b^2}$$ 15. Therefore, $$F(s) = 2 \times \frac{s + a}{(s + a)^2 + b^2} + \frac{16}{5b} \times \frac{b}{(s + a)^2 + b^2}$$ 16. Taking inverse Laplace transform: $$f(t) = 2 e^{-at} \cos(bt) + \frac{16}{5b} e^{-at} \sin(bt)$$ 17. Substitute back $a = \frac{7}{10}$ and $b = \frac{\sqrt{191}}{10}$: $$f(t) = 2 e^{-\frac{7}{10}t} \cos\left(\frac{\sqrt{191}}{10} t\right) + \frac{16}{5 \times \frac{\sqrt{191}}{10}} e^{-\frac{7}{10}t} \sin\left(\frac{\sqrt{191}}{10} t\right)$$ 18. Simplify the coefficient: $$\frac{16}{5 \times \frac{\sqrt{191}}{10}} = \frac{16 \times 10}{5 \sqrt{191}} = \frac{160}{5 \sqrt{191}} = \frac{32}{\sqrt{191}}$$ 19. Final answer: $$f(t) = 2 e^{-\frac{7}{10}t} \cos\left(\frac{\sqrt{191}}{10} t\right) + \frac{32}{\sqrt{191}} e^{-\frac{7}{10}t} \sin\left(\frac{\sqrt{191}}{10} t\right)$$