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

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Answer 1 · 2017-11-09T23:20:57

Expand the brackets and use the Formula

$x=1/4(13+sqrt95i)$ or $x=1/4(13-sqrt95i)$

Firstly, expand the brackets and collect like terms.

$y=-(x^2-12x+36)-x^2+x+3$
$y=-x^2+12x-36-x^2+x+3$
$y=-2x^2+13x-33$

Now we sub into the quadratic formula $a=-2, b=13, c=-33$

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

$x=(-13+-sqrt(13^2-4(-2)(-33)))/(2(-2)$

$x=(-13+-sqrt(169-264))/(-4)$

$x=(13+-sqrt(-95))/4$

Ok, we can see here our roots will be imaginary, but that's ok.

$x=(13+-sqrt(95)i)/(4)$

$x=1/4(13+sqrt95i)$ or $x=1/4(13-sqrt95i)$

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