# Question #11a9a

Feb 6, 2017

The discriminant is negative, so no real solutions exist.

The imaginary solutions are $x = - 1.5 \pm \frac{\sqrt{- 11}}{2}$

#### Explanation:

Since the equation is already in the standard form $a {x}^{2} + b x + c = 0$, the easiest way is to use the quadratic formula.

First, test the discriminant: $\sqrt{{b}^{2} - 4 a c}$

This is negative: $\sqrt{9 - 20} = \sqrt{- 11}$

This means there are no real solutions.

But, continuing (in case you are looking for imaginary solutions)

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

$x = \frac{\left(- 3\right) \pm \sqrt{{3}^{2} - 4 \left(1\right) \left(5\right)}}{2 \left(1\right)}$

$x = \frac{\left(- 3\right) \pm \sqrt{9 - 20}}{2}$

$x = - 1.5 \pm \frac{\sqrt{- 11}}{2}$