How do you find the roots, real and imaginary, of $y= x^2 - 5x + 6 - (2x-1)^2$ using the quadratic formula?

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Answer 1 · 2018-06-09T09:27:50

$color(crimson)(x = -((1 + sqrt61) / 6), -((1 - sqrt61) / 6)$

$y = x^2 - 5x + 6 - 4x^2 + 4x - 1$

$y = -3x^2 -x + 5$

$a = -3, b = -1, c = 5$

$x = (-b +- sqrt(b^2 - 4ac)) / (2a) " is the quadratic formula"$

$:. x = (1 +- sqrt(1 + 60)) / -6$

$color(crimson)(x = -((1 + sqrt61) / 6), -((1 - sqrt61) / 6)$

How do you find the roots, real and imaginary, of y= x^2 - 5x + 6 - (2x-1)^2 y= x^2 - 5x + 6 - (2x-1)^2 using the quadratic formula?