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

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