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

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Answer 1 · 2017-04-18T05:38:20

Roots are $(5+isqrt71)4$ and $(5-isqrt71)4$

$y=-x^2-x-3-(x-3)^2$

= $-x^2-x-3-(x^2-6x+9)$

= $-x^2-x-3-x^2+6x-9$

= $-2x^2+5x-12$

Using quadratic formula, the roots are $(-5+-sqrt(5^2-4×(-2)×(-12)))/(2×(-2)$

= $(-5+-sqrt(25-96))/(-4)$

= $(5+-sqrt(-71))/4$

i.e. roots are $(5+isqrt71)4$ and $(5-isqrt71)4$

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