How do you find the roots, real and imaginary, of #y=3(x - 6)^2-x + 3# using the quadratic formula?

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Answer 1 · 2016-02-13T06:47:42

Roots are #(37+sqrt37)/6# and #(37-sqrt37)/6#, both real but irrational

Converting the expression to its general form

#3(x−6)^2−x+3=3(x^2-12x+36)-x+3#

or #3x^2-37x+111#

Quadratic formula tell that the root of this are given by

#(-b+-sqrt(b^2-4ac))/(2a)#

As #a=3, b=-37 and c=111#, roots of given equation are given by

#(37+-sqrt(1369-1332))/6# i.e. #(37+-sqrt37)/6#

i.e. #(37+sqrt37)/6# and #(37-sqrt37)/6#, both real but irrational