Question

In the following exercises, solve by using the Quadratic Formula.
$\frac{1}{3}{n}^2+n=-\frac{1}{2}$

Short answer

The solutions of the given equation are $n=-\frac{3}{2}\pm \frac{\sqrt{3}}{2}$.

Step-by-step solution

Step 1. Write the quadratic equation in standard form.

The given equation is:

$\frac{1}{3}{n}^2+n=-\frac{1}{2}$

The given equation can be written as:

$\begin{array}{rcl}\frac{1}{3}{n}^2+n+\frac{1}{2}& =& 0\\ \frac{2n^2+6n+3}{6}& =& 0\\ 2n^2+6n+3& =& 0\end{array}$

Step 2. Substitute the values in Quadratic Formula.

Quadratic Formula for the equation $a{x}^2+bx+c=0$is:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

In the given equation, $a=2,b=6,c=3$. Substitute these values in Quadratic Formula.

$n=\frac{-6\pm \sqrt{(6{)}^2-4(2\left)\right(3)}}{2\left(2\right)}$

Step 3. Simplify the value.

On simplification, we get

$\begin{array}{rcl}n& =& \frac{-6\pm \sqrt{36-24}}{4}\\ n& =& \frac{-6\pm \sqrt{12}}{4}\\ n& =& \frac{-6}{4}\pm \frac{2\sqrt{3}}{4}\\ n& =& -\frac{3}{2}\pm \frac{\sqrt{3}}{2}\end{array}$

Step 4. Conclusion.

The solutions of the given equation are $n=-\frac{3}{2}\pm \frac{\sqrt{3}}{2}$.