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

Question

1369 views around the world

You can reuse this answer
Creative Commons License

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#