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

Nov 16, 2016

#### Explanation:

Multiply both sides by 14:

$4 {c}^{2} - 7 c - 3 = 0$

Check the discriminant

${b}^{2} - 4 \left(a\right) \left(c\right) = {\left(- 7\right)}^{2} - 4 \left(4\right) \left(- 3\right) = 97$

This does not have complex roots; only real roots.

$c = \frac{- b \pm \sqrt{{b}^{2} - 4 \left(a\right) \left(c\right)}}{2 a}$

$c = \frac{7 + \sqrt{97}}{8} \mathmr{and} c = \frac{7 - \sqrt{97}}{8}$