Question 369495
f(x)={{{ 4x^2-4x+1 = 0}}}
{{{(2x-1)^2 = 0}}}
(2x-1)*(2x-1) = 0
2x = 1
x = 1/2 is the answer, multiplicity 2
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There are 2 answers, both of them are x = 1/2
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*[invoke solve_quadratic_equation 4,-4,1]
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Ignore the comments about 1 solution.  There's one value for the 2 solutions.