How do you find the roots, real and imaginary, of $y=(x+6)(-2x+5)$ using the quadratic formula?

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Answer 1 · 2016-03-14T12:04:06

There are two real roots: $x=-6$ and $x=5/2$ . There are no imaginary roots.

First, expand the original equation to $y=ax^2+bx+c$ form.

$y=-2x^2+5x-12x+30=-2x^2-7x+30$

Next plug the coefficients into the quadratic formula, giving you

$x=(7+-sqrt(49-4(-2)(30)))/(2(-2))$
$=(7+-sqrt(289))/(-4)$
$=(7+-17)/(-4)$

So

$x=24/-4=-6$ or $x=(-10)/-4=5/2$

How do you find the roots, real and imaginary, of y=(x+6)(-2x+5)y=(x+6)(-2x+5) using the quadratic formula?