From 6/√-36-12m) = (11-t)/(6-m), 36-6m = (1 1-t) √-36.

① Substitute ② to get 36-t? - 1 1t+36﹚=( 1 1-t)√[36-2﹙t？ - 1 1t+36﹚]

﹣t(t- 1 1)=﹣(t- 1 1)√[-2﹙t？ - 1 1t+ 18﹚]

﹣t(t- 1 1)+(t- 1 1)√[-2﹙t？ - 1 1t+ 18﹚]=0

﹣(t- 1 1)﹛t-√[-2﹙t？ - 1 1t+ 18﹚]﹜=0

-(t- 1 1) = 0 or t-√ [-2-t? - 1 1t+ 18﹚]=0

For t-√ [-2] t? -11t+18] = 0, which means √ [-2] t? - 1 1t+ 18﹚]=t(2≦t≦9)

-2﹙t？ - 1 1t+ 18﹚=t？ That's -2t? +22t-36=t？

3t？ -22t+36=0

Δ=﹙﹣22﹚? -4×3×36=52>0

t=﹙22 √52﹚/6=﹙ 1 1√ 13)/3(2≤t≤9)

To sum up, t = 1 1, ﹙1+√13)/3, ﹙1√/kloc.