Question 82585
Good job!! I agree with your answer. You correctly initially squared both sides of the equation in order to get rid of the radical; used FOIL on the left side; and solved for the x-term. 
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If your instructor/book wanted you to solve for the x=term, just continue the process:
  0=x^2-7x+10
(x-5)(x-2)=0
x-5=0
x=5
or x-2=0
x=2
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Try plugging each x-value back into the original equation to make sure each solution is a true solution.  One of the values of x could possibly be an erroneous solution:
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{{{(sqrt(x-1))}}}={{{(x-3)}}}
Let x=5
{{{(sqrt(5-1))}}}={{{(5-3)}}}
{{{(sqrt(4))}}}={{{(2)}}}[the square root of 4 is either a plus or minus 2]
2=2 [so, x=5 checks out]
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{{{(sqrt(x-1))}}}={{{(x-3)}}}
Let x=2
{{{(sqrt(-1))}}}={{{(-1)}}} [the square root of 1 is either a plus or minus (1).  In this case, the (-1) works]

-1 = -1 [ture, so x=2 is also a solution]