Question

$g(2) = 13$

$$f(g(2)) = f(13) $$Now $4\text{ }g(f(x)) = 3x^{2} - 32x + 72$

$$4\left\lbrack 3f^{2}(x) + 2f(x) - 3 \right\rbrack = 3x^{2} - 32x + 72 $$Let $f(x) = t$

$${12t^{2} + 8t - \left( 3x^{2} - 32x + 84 \right) = 0 }{f(x) = \frac{- 8 \pm \sqrt{64 + 48\left( 3x^{2} - 32x + 84 \right)}}{24} }{f(x) = \frac{- 8 \pm 4(3x - 16)}{24} }$$$\because f(0) = - 3\ \therefore$ we take $+ ve$ sign

$${\therefore f(x) = \frac{- 8 + 4(3x - 16)}{24} }{\therefore f(13) = \frac{- 8 + 4 \times 23}{24} = \frac{84}{24} = \frac{7}{2}}$$