How do you find the zeros, real and imaginary, of $y=-6x^2-12x+9$ using the quadratic formula?

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Answer 1 · 2017-07-31T12:13:56

$x=(-2+-sqrt10)/2$

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

Here, $a=-6$ , $b=-12$ , $c=9$
So,

$x=(12+-sqrt(144+216))/(-12)$
$rArr$ $x=(-2+-sqrt10)/2$

Or in decimal form,

$x=0.58113883008418966599944677221636$
or, $x=-2.5811388300841896659994467722164$

How do you find the zeros, real and imaginary, of y=-6x^2-12x+9y=-6x^2-12x+9 using the quadratic formula?