\( \left(x^{2}-6 x\right)^{2}+13\left(x^{2}-6 x\right)=-40 \) The solution set is \( \{\square\} \)

Let's start by making a substitution to make the problem easier to handle. Let \( y = x^2 - 6x \). The equation now transforms to: \[ y^2 + 13y + 40 = 0. \] This is a standard quadratic equation and we can solve it using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). In our case, \( a = 1 \), \( b = 13 \), and \( c = 40 \). Calculating the discriminant: \[ b^2 - 4ac = 13^2 - 4 \cdot 1 \cdot 40 = 169 - 160 = 9. \] Thus, the roots are: \[ y = \frac{-13 \pm \sqrt{9}}{2 \cdot 1} = \frac{-13 \pm 3}{2}. \] Calculating the two potential values for \( y \): 1. \( y = \frac{-13 + 3}{2} = \frac{-10}{2} = -5 \) 2. \( y = \frac{-13 - 3}{2} = \frac{-16}{2} = -8 \) Now, we substitute back to \( x^2 - 6x = y \). For \( y = -5 \): \[ x^2 - 6x + 5 = 0, \] which can be solved using the quadratic formula: \[ x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} = \frac{6 \pm \sqrt{36 - 20}}{2} = \frac{6 \pm \sqrt{16}}{2} = \frac{6 \pm 4}{2}. \] This gives us: 1. \( x = \frac{10}{2} = 5 \) 2. \( x = \frac{2}{2} = 1 \) For \( y = -8 \): \[ x^2 - 6x + 8 = 0, \] Using the quadratic formula here as well: \[ x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1} = \frac{6 \pm \sqrt{36 - 32}}{2} = \frac{6 \pm \sqrt{4}}{2} = \frac{6 \pm 2}{2}. \] This provides: 1. \( x = \frac{8}{2} = 4 \) 2. \( x = \frac{4}{2} = 2 \) In total, we have \( x = 1, 2, 4, 5 \). Therefore, the solution set is: \(\{1, 2, 4, 5\}\)