How do you solve the quadratic using the quadratic formula given #-c^2-6c+8=0# over the set of complex numbers?

Question

1086 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-02T12:45:14

#c=(-6+-2sqrt(17))/2#

The quadratic formula states that if you have an equation in the form:
#nx^2+px+q=0#

The solution(s) will be:
#x=(-p+-sqrt(p^2-4nq))/(2n)#

In our case #n=-1,p=-6# and #q=8#, and applying the quadratic formula gives us:
#c=(-(-6)+-sqrt((-6)^2-4*-1*8))/(2*-1)#

#c=(6+-sqrt(36-(-32)))/-2#

#c=-(6+-sqrt(36+32))/2#

#c=(-6+-sqrt(68))/2#

#c=(-6+-sqrt(2*2*17))/2#

#c=(-6+-2sqrt(17))/2#