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

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Answer 1 · 2016-06-08T13:37:51

$(-7+-sqrt41)/2$

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

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

= $-x^2-7x-2$

As according to quadratic formula roots of $ax^2+bx+c=0$ are

$(-b+-sqrt(b^2-4ac))/(2a)$ , roots of $-x^2-7x-2$ are

$(-(-7)+-sqrt((-7)^2-4(-1)(-2)))/(2(-1))$

= $(7+-sqrt(49-8))/(-2)$

= $(-7+-sqrt41)/2$

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