# How do you solve the quadratic using the quadratic formula given b^2-7b-3=0 over the set of complex numbers?

$x = \frac{7}{2} \pm \frac{\sqrt{61}}{2}$ The answer is not complex
Quadratic formula when using $A {x}^{2} + B x + C = 0$:
$x = \frac{- B \pm \sqrt{{B}^{2} - 4 A C}}{2 A}$
$x = \frac{7 \pm \sqrt{49 - 4 \left(1\right) \left(- 3\right)}}{2 \left(1\right)} = \frac{7 \pm \sqrt{61}}{2} = \frac{7}{2} \pm \frac{\sqrt{61}}{2}$