How do you find the roots, real and imaginary, of $y=-x^2-12x -7$ using the quadratic formula?

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Answer 1 · 2016-06-07T11:50:23

$12+-sqrt(29)$ . There are no imaginary roots.

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

and $a=-1, b= -12, c=-7$

Now let's substitute in:

$x=(12+-sqrt((-12)^2-4(-1)(-7)))/(2(-1))$

and now simplify:

$x=12+-sqrt((144-28))/-2$

$x=12+-sqrt(116)/-2=12+-(2sqrt(29))/-2=12+-sqrt(29)$

There are no imaginary roots. To explicitly state there are no imaginary roots, we can write the answer this way:

$x=12+sqrt(29)+0i, 12-sqrt(29)+0i$

How do you find the roots, real and imaginary, of y=-x^2-12x -7y=-x^2-12x -7 using the quadratic formula?