(х-5)⁴+(х+5)²-12=0

Let's solve the equation \((x - 5)^4 + (x + 5)^2 - 12 = 0\). 1. **Introduce substitution if needed**: Let \( y = (x - 5)^2 \). Then \((x + 5)^2 = (x - 5 + 10)^2 = (y + 10)^2\), and thus we can rewrite the equation: \[ y^2 + (y + 10)^2 - 12 = 0 \] 2. **Expand and simplify**: \[ (y + 10)^2 = y^2 + 20y + 100 \] We can substitute that back into the equation: \[ y^2 + (y^2 + 20y + 100) - 12 = 0 \] Combine like terms: \[ 2y^2 + 20y + 100 - 12 = 0 \\ 2y^2 + 20y + 88 = 0 \] 3. **Divide the entire equation by 2 to simplify**: \[ y^2 + 10y + 44 = 0 \] 4. **Use the quadratic formula**: The quadratic formula is given by: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1 \), \( b = 10 \), and \( c = 44 \): \[ y = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 1 \cdot 44}}{2 \cdot 1} \\ y = \frac{-10 \pm \sqrt{100 - 176}}{2} \\ y = \frac{-10 \pm \sqrt{-76}}{2} \\ y = \frac{-10 \pm \sqrt{76}i}{2} \\ y = -5 \pm \sqrt{19}i \] 5. **These values of \( y \) correspond to \((x - 5)^2\)**: Now we have: \[ (x - 5)^2 = -5 + \sqrt{19}i \] and \[ (x - 5)^2 = -5 - \sqrt{19}i \] 6. **Solve for \( x \)**: For the first equation: \[ x - 5 = \pm \sqrt{-5 + \sqrt{19}i} \\ x = 5 \pm \sqrt{-5 + \sqrt{19}i} \] For the second equation: \[ x - 5 = \pm \sqrt{-5 - \sqrt{19}i} \\ x = 5 \pm \sqrt{-5 - \sqrt{19}i} \] Thus, we find the solutions for \( x \) in terms of complex numbers. The equation has no real solutions due to the presence of the complex numbers.