How do you find the roots, real and imaginary, of $y=-2(x+1)^2-(x-3)^2-10$ using the quadratic formula?

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Answer 1 · 2018-06-09T09:10:43

$color(blue)(x = (1 - isqrt62)/3, (1 + i sqrt62) / 3$

$y = -2(x+1)^2 - (x-3)^2 - 10$

$y = -2x^2 - 4x - 2 -x^2 + 6x - 9 - 10$

$y = -3x^2 + 2x - 21$

$a = -3, b = 2, c = =-21$

$x = (-b +- sqrt(b^2 - 4ac)) / (2a)$

$x = (-2 +- sqrt(4 - 252)) / - 6$

$color(blue)(x = (1 - isqrt62)/3, (1 + i sqrt62) / 3$

How do you find the roots, real and imaginary, of y=-2(x+1)^2-(x-3)^2-10 y=-2(x+1)^2-(x-3)^2-10 using the quadratic formula?