How do you solve the quadratic with complex numbers given #2/7c^2-1/2c-3/14=0#?

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Answer 1 · 2016-11-16T01:49:17

Please see the explanation.

Multiply both sides by 14:

#4c^2 - 7c - 3 = 0#

Check the discriminant

#b^2 - 4(a)(c) = (-7)^2 - 4(4)(-3) = 97#

This does not have complex roots; only real roots.

#c = (-b +-sqrt(b^2 - 4(a)(c)))/(2a)#

#c = (7 + sqrt(97))/8 and c = (7 - sqrt(97))/8#