Find all real and imaginary solutions to the equation. \[ \left(x^{2}-6 x\right)^{2}-35\left(x^{2}-6 x\right)-200=0 \]

Let's define a substitution to simplify the equation. Let: \[ y = x^2 - 6x \] Substituting \( y \) into the given equation, we have: \[ (y^2 - 35y - 200 = 0) \] Now, this is a standard quadratic equation in \( y \). We can solve it using the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1 \), \( b = -35 \), and \( c = -200 \). Plugging these values into the formula, we get: \[ y = \frac{35 \pm \sqrt{(-35)^2 - 4 \cdot 1 \cdot (-200)}}{2 \cdot 1} \] Calculating the discriminant: \[ (-35)^2 = 1225 \] \[ -4 \cdot 1 \cdot (-200) = 800 \] Thus, we have: \[ b^2 - 4ac = 1225 + 800 = 2025 \] Taking the square root of the discriminant: \[ \sqrt{2025} = 45 \] Now, substituting back into the quadratic formula: \[ y = \frac{35 \pm 45}{2} \] Calculating the two possible values for \( y \): 1. \( y = \frac{80}{2} = 40 \) 2. \( y = \frac{-10}{2} = -5 \) Now, we have two cases for \( y \): \( y = 40 \) and \( y = -5 \). **Case 1:** \( y = 40 \) From our substitution, we have: \[ x^2 - 6x = 40 \] \[ x^2 - 6x - 40 = 0 \] Applying the quadratic formula again with \( a = 1 \), \( b = -6 \), and \( c = -40 \): \[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot (-40)}}{2 \cdot 1} \] \[ = \frac{6 \pm \sqrt{36 + 160}}{2} \] \[ = \frac{6 \pm \sqrt{196}}{2} \] \[ = \frac{6 \pm 14}{2} \] Calculating the two potential solutions: 1. \( x = \frac{20}{2} = 10 \) 2. \( x = \frac{-8}{2} = -4 \) **Case 2:** \( y = -5 \) Now we solve: \[ x^2 - 6x = -5 \] \[ x^2 - 6x + 5 = 0 \] Again applying the quadratic formula with \( a = 1 \), \( b = -6 \), and \( c = 5 \): \[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} \] \[ = \frac{6 \pm \sqrt{36 - 20}}{2} \] \[ = \frac{6 \pm \sqrt{16}}{2} \] \[ = \frac{6 \pm 4}{2} \] Again, calculating the two potential solutions: 1. \( x = \frac{10}{2} = 5 \) 2. \( x = \frac{2}{2} = 1 \) Combining all the solutions, the real solutions to the initial equation are: \[ x = 10, \quad x = -4, \quad x = 5, \quad x = 1 \] Thus, all solutions (real and imaginary) are: **Real Solutions:** \[ \boxed{10, -4, 5, 1} \] **Imaginary Solutions:** There are no imaginary solutions since all derived solutions for \( x \) are real.