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

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Answer 1 · 2018-02-19T21:47:04

The roots are:
$(5+sqrt(17))/(-4), and(5-sqrt(17))/(-4)$

First, expand the equation:

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

$-2x^2-4x-2-x+1$

$-2x^2-5x-1$

Now that it is in $ax^2+bx+c$ standard form, with

$a=-2,b=-5,c=-1$ ,

you can plug into the quadratic formula:

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

$(-(-5)+-sqrt((-5)^2-4*-2*-1))/(2*-2)$

$(5+-sqrt(17))/(-4)$

The answer is

$(5+sqrt(17))/(-4), and(5-sqrt(17))/(-4)$

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