In bridge each of the Four players (A. B. C. D) receives 13 Cards Suppose A and C have 11 of the 13 Spades between Them What is the probability that the remaining two spades are distributed so that B and and D have one spade apiece? ​

i can you with this answer

step-by-step explanation:

27 +27 or 27times 2=54

the diffrence is 0

"standard form:

3x2 + 14x − 10 = 0

solutions based on quadratic formula:

x1   = −14 − √ 142 − 4×3×(−10) /   2×3   = −14 − 2 × √ 79 /   6   ≈ −5.296

x2   = −14 + √ 142 − 4×3×(−10) /   2×3   = −14 + 2 × √ 79 /   6   ≈ 0.629