How do you find the roots, real and imaginary, of $y= 2x^2 - 2x - (x-9)^2$ using the quadratic formula?

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Answer 1 · 2018-07-14T20:17:19

$x=-8pmsqrt(145)$

Simplifying the given equation and using the Formula
$(a-b)^2=a^2-2ab+b^2$

we get
$2x^2-2x-(x^2-18x+81)=0$ combinging like Terms

$x^2+16x-81=0$

$x^2+px+q=0$

$x_(1,2)=-p/2pmsqrt((p/2)^2-q)$

we get

$x_(1,2)=-8pmsqrt(64+81)$

$x_(1,2)=-8pmsqrt(145)$

How do you find the roots, real and imaginary, of y= 2x^2 - 2x - (x-9)^2 y= 2x^2 - 2x - (x-9)^2 using the quadratic formula?