How do you find the roots, real and imaginary, of $y= -3x^2 -+5x-2$ using the quadratic formula?

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Answer 1 · 2018-02-04T08:43:03

$x_1=6/(-6)=-1$

$x_2=4/(-6)=-2/3$

The quadratic formula states that if you have a quadratic in the form $ax^2+bx+c=0$ , the solutions are:
$x=(-b+-sqrt(b^2-4ac))/(2a)$

In this case, $a=-3$ , $b=-5$ and $c=-2$ . We can plug this into the quadratic formula to get:
$x=(-(-5)+-sqrt((-5)^2-4*-3*-2))/(2*-3)$

$x=(5+-sqrt(25-24))/(-6)=(5+-1)/(-6)$

$x_1=6/(-6)=-1$

$x_2=4/(-6)=-2/3$

How do you find the roots, real and imaginary, of y= -3x^2 -+5x-2y= -3x^2 -+5x-2 using the quadratic formula?