Question 167340
Your method is called "completing the square"...
This site explains it quite nicely:
http://www.purplemath.com/modules/sqrquad.htm
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x^2-2x-13=0
x^2-2x = 13
Take half the 'b' coefficient and square it:[(1/2)(-2)]^2 = [-1]^1 = 1 
x^2-2x+1 = 13+1  (since you added 1 to left, do so on the right - for balance)
(x-1)^2 = 14
x-1 = sqrt(14)
x = 1(+-)sqrt(14)
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That's 1 "plus or minus" square root of 14.
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4x^2-4x+3=0
4x^2-4x = -3
factor the 4 on the left:
4(x^2-x) = -3
(x^2-x) = -3/4
(x^2-x+(1/4)) = -3/4 + 1/4
(x-(1/2))^2 = -2/4
x-(1/2) = sqrt(-2/4)
x = (1/2)(+-)sqrt(-1/2)
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Since the term inside the sqrt is negative -- we have no real solutions -- rather, we have two imaginary solutions.
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You could see the same thing using the "quadratic equation" below:
*[invoke quadratic "x", 4, -4, 3 ]