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

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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:
$c=(-(-6)+-sqrt((-6)^2-4*-1*8))/(2*-1)$

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

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

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

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

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

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