$\mathcal{L}(5t - 2) = 5\mathcal{L}(t) - 2\mathcal{L}(1)$
$\mathcal{L}(5t - 2) = 5\left( \dfrac{1}{s^2} \right) - 2\left( \dfrac{1}{s} \right)$
$\mathcal{L}(5t - 2) = \dfrac{5}{s^2} - \dfrac{2}{s}$
$\mathcal{L}(5t - 2) = \dfrac{1}{s^2}(5 - 2s)$ answer