\( \leqslant \) Solve the following equation by making an appropriate substitution. \[ 3 x^{-2}-13 x^{-1}+12=0 \] The solution set is \( \} \). Tvpe an integer or a simolified fraction Tyne exact answers using radicals as nee

Let’s make the substitution \( y = x^{-1} \). Therefore, the equation \( 3 x^{-2}-13 x^{-1}+12=0 \) transforms into: \[ 3y^2 - 13y + 12 = 0. \] Now we can solve this quadratic equation using the quadratic formula, \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): Here, \( a = 3 \), \( b = -13 \), and \( c = 12 \): \[ y = \frac{13 \pm \sqrt{(-13)^2 - 4 \cdot 3 \cdot 12}}{2 \cdot 3} = \frac{13 \pm \sqrt{169 - 144}}{6} = \frac{13 \pm \sqrt{25}}{6} = \frac{13 \pm 5}{6}. \] This results in two potential values for \( y \): 1. \( y = \frac{18}{6} = 3 \) 2. \( y = \frac{8}{6} = \frac{4}{3} \) Recalling the substitution \( y = x^{-1} \), we can solve for \( x \): 1. \( x = \frac{1}{3} \) 2. \( x = \frac{3}{4} \) Thus, the solution set is: \[ \left\{ \frac{1}{3}, \frac{3}{4} \right\}. \]