How do you find the roots, real and imaginary, of $y = (- 2 x - 4) (x + 1) + 3 x^{2} + 4 x$ $y=(-2x-4)(x+1)+3x^2+4x$ using the quadratic formula?

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Answer 1 · 2018-04-03T12:47:17

$x = 2 \pm \sqrt{5}$ $x=2+-sqrt5$

When expanded,

$y = x^{2} - 2 x - 4$ $y=x^2-2x-4$

$x = \frac{2 \pm \sqrt{4 + 16}}{2}$ $x=(2+-sqrt(4+16))/2$
$x = 2 \pm \frac{\sqrt{20}}{2}$ $x=2+-sqrt20/2$
$x = 2 \pm \sqrt{5}$ $x=2+-sqrt5$

How do you find the roots, real and imaginary, of y=(-2x-4)(x+1)+3x^2+4x y=(−2x−4)(x+1)+3x2+4xy=(-2x-4)(x+1)+3x^2+4x using the quadratic formula?