Solve the equation. \[ 4(x+1)^{4}-53(x+1)^{2}=-49 \]

To solve the equation \[ 4(x+1)^{4} - 53(x+1)^{2} = -49, \] let's make a substitution. Set \( y = (x+1)^{2} \). Then, the equation transforms into: \[ 4y^{2} - 53y + 49 = 0. \] Now, we can solve this quadratic equation using the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \] where \( a = 4 \), \( b = -53 \), and \( c = 49 \). Calculating the discriminant: \[ b^2 - 4ac = (-53)^{2} - 4 \cdot 4 \cdot 49 = 2809 - 784 = 2025. \] Now, we find the square root of the discriminant: \[ \sqrt{2025} = 45. \] Now substituting back into the quadratic formula: \[ y = \frac{53 \pm 45}{2 \cdot 4} = \frac{53 \pm 45}{8}. \] This gives us two solutions for \( y \): 1. \( y = \frac{98}{8} = \frac{49}{4} \) 2. \( y = \frac{8}{8} = 1 \) Now, recall that \( y = (x + 1)^{2} \). We need to solve for \( x \) in both cases. **Case 1:** \[ (x + 1)^{2} = \frac{49}{4}. \] Taking the square root: \[ x + 1 = \pm \frac{7}{2}. \] Thus, we have two solutions: \[ x = \frac{7}{2} - 1 = \frac{5}{2} \quad \text{and} \quad x = -\frac{7}{2} - 1 = -\frac{9}{2}. \] **Case 2:** \[ (x + 1)^{2} = 1. \] Taking the square root: \[ x + 1 = \pm 1. \] Thus, we have two more solutions: \[ x = 1 - 1 = 0 \quad \text{and} \quad x = -1 - 1 = -2. \] Now, combining all solutions, we find: \[ x = \frac{5}{2}, \quad x = -\frac{9}{2}, \quad x = 0, \quad x = -2. \] So the final solutions to the original equation are: \[ \boxed{\left\{ \frac{5}{2}, -\frac{9}{2}, 0, -2 \right\}}. \]