Help!

First we will need to find g(x).

$g(3x-8) = 2x^2+5x-3\\ \quad g(x) \\= g\left(3\left(\dfrac{x+8}{3}\right)-8\right) \\= 2\left(\dfrac{x+8}{3}\right)^2+5\left(\dfrac{x+8}{3}\right)-3\\ =\dfrac{2}{9}\left(x^2+16x+64\right)+\dfrac{5}{3}(x+8)-3\\ =\dfrac{2}{9}x^2 +\dfrac{47}{9}x+\dfrac{221}{9}$

Therefore g(-5) = $\dfrac{2}{9}(-5)^2+\dfrac{47}{9}(-5)+\dfrac{221}{9}=\dfrac{50-235+221}{9} = \dfrac{36}{9} = 4$

Ans = 4.