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

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Answer 1 · 2017-11-10T14:43:03

$:. x = (-1 + sqrt33) / 4 ~= 1.186$

$:. x = (-1 - sqrt33) / 4 ~= -1.686$

There are no imaginary roots.

Let's first simplify:

$y=(3x-4)(x+2)-x^2-x$

$y=3x^2+2x-8-x^2-x$

$y=2x^2+x-8$

$x = (-b \pm sqrt(b^2-4ac)) / (2a)$

with $a=2, b=1, c=-8$

$x = (-1 \pm sqrt(1^2-4(2)(-8))) / (2(2))$

$x = (-1 \pm sqrt(1^2+32)) / 4$

$x = (-1 \pm sqrt33) / 4$

$:. x = (-1 + sqrt33) / 4 ~= 1.186$

$:. x = (-1 - sqrt33) / 4 ~= -1.686$

There are no imaginary roots.

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