1. The problem is to solve the equation $$\frac{t}{t+16} + \frac{t+16}{t} = \frac{2 \times 1}{12} = \frac{1}{6}.$$\n\n2. Let's rewrite the equation clearly:\n$$\frac{t}{t+16} + \frac{t+16}{t} = \frac{1}{6}.$$\n\n3. To combine the left side, find a common denominator $t(t+16)$:\n$$\frac{t^2}{t(t+16)} + \frac{(t+16)^2}{t(t+16)} = \frac{1}{6}.$$\n\n4. Combine the fractions:\n$$\frac{t^2 + (t+16)^2}{t(t+16)} = \frac{1}{6}.$$\n\n5. Expand $(t+16)^2$:\n$$t^2 + (t^2 + 32t + 256) = 2t^2 + 32t + 256.$$\n\n6. Substitute back into numerator:\n$$\frac{2t^2 + 32t + 256}{t(t+16)} = \frac{1}{6}.$$\n\n7. Cross multiply to solve for $t$:\n$$6(2t^2 + 32t + 256) = t(t+16).$$\n\n8. Expand both sides:\nLeft side: $12t^2 + 192t + 1536$\nRight side: $t^2 + 16t$\n\n9. Bring all terms to one side:\n$$12t^2 + 192t + 1536 - t^2 -16t = 0,$$\nwhich simplifies to:\n$$11t^2 + 176t + 1536 = 0.$$\n\n10. Simplify by dividing all terms by 11:\n$$t^2 + 16t + \frac{1536}{11} = 0.$$\n\n11. Use the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1$, $b=16$, and $c=\frac{1536}{11}$:\n$$t = \frac{-16 \pm \sqrt{16^2 - 4 \times 1 \times \frac{1536}{11}}}{2} = \frac{-16 \pm \sqrt{256 - \frac{6144}{11}}}{2}.$$\n\n12. Compute under the square root:\n$$256 = \frac{2816}{11},$$\nso\n$$\sqrt{\frac{2816}{11} - \frac{6144}{11}} = \sqrt{-\frac{3328}{11}}.$$\n\n13. Since the discriminant is negative, there are no real solutions; the roots are complex.\n\n14. Final answer: $$t = \frac{-16 \pm i\sqrt{\frac{3328}{11}}}{2} = -8 \pm i \frac{\sqrt{3328/11}}{1}.$$\n\nSimplify $\sqrt{3328/11}$ if desired, or leave as is.\n\nTherefore, the equation has two complex solutions and no real solutions.