Solve \( t^{4}-20 t^{2}+64=0 \) List them separated by commas,

To solve the equation \( t^{4}-20t^{2}+64=0 \), we can make a substitution. Let \( u = t^{2} \). Then the equation becomes: \[ u^{2} - 20u + 64 = 0 \] Now, we can apply the quadratic formula: \[ u = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \] Here, \( a = 1, b = -20, \) and \( c = 64 \): \[ u = \frac{20 \pm \sqrt{(-20)^{2} - 4 \cdot 1 \cdot 64}}{2 \cdot 1} \] \[ = \frac{20 \pm \sqrt{400 - 256}}{2} \] \[ = \frac{20 \pm \sqrt{144}}{2} \] \[ = \frac{20 \pm 12}{2} \] This gives us two values for \( u \): 1. \( u = \frac{32}{2} = 16 \) 2. \( u = \frac{8}{2} = 4 \) Now, substituting back \( u = t^{2} \): 1. For \( t^{2} = 16 \): \( t = 4 \) or \( t = -4 \) 2. For \( t^{2} = 4 \): \( t = 2 \) or \( t = -2 \) Therefore, the solutions for the original equation are: \( t = 4, -4, 2, -2 \)