How do you find the roots, real and imaginary, of $y=-2x^2+6x-(2x-7)^2$ using the quadratic formula?

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Answer 1 · 2018-01-29T15:44:49

$x=(17+isqrt(-5))/6,(17-isqrt(-5))/6$

First, expand the equation.

$-2x^2+6x-(4x^2-28x+49)$

$-2x^2-4x^2+6x+28x-49$

$-6x^2+34x-49$

$6x^2-34+49$

The quadratic equation is $x=(-b+-sqrt(b^2-4ac))/(2a)$ .

Here, $a=6,b=-34,c=49$

Input:

$(-(-34)+-sqrt((-34)^2-4*6*49))/(2*6)$

$(34+-sqrt(1156-1176))/12$

$(34+-sqrt(-20))/12$

$(34+sqrt(-20))/12,(34-sqrt(-20))/12$

$(34+2isqrt(-5))/12,(34-2isqrt(-5))/12$

$x=(17+isqrt(-5))/6,(17-isqrt(-5))/6$

How do you find the roots, real and imaginary, of y=-2x^2+6x-(2x-7)^2 y=-2x^2+6x-(2x-7)^2 using the quadratic formula?