Find the Exact Value (-8+ square root of -12)/40

$\frac{- 8 + \sqrt{- 12}}{40}$

Pull out imaginary unit $i$ .

Rewrite $- 12$ as $- 1 (12)$ .

$\frac{- 8 + \sqrt{- 1 (12)}}{40}$

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$\frac{- 8 + \sqrt{- 1} \cdot \sqrt{12}}{40}$

Rewrite $\sqrt{- 1}$ as $i$ .

$\frac{- 8 + i \cdot \sqrt{12}}{40}$

Simplify the numerator.

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$\frac{- 8 + i \cdot \sqrt{4 (3)}}{40}$

Rewrite $4$ as $2^{2}$ .

$\frac{- 8 + i \cdot \sqrt{2^{2} \cdot 3}}{40}$

Pull terms out from under the radical.

$\frac{- 8 + i \cdot (2 \sqrt{3})}{40}$

Move $2$ to the left of $i$ .

$\frac{- 8 + 2 i \sqrt{3}}{40}$

Simplify terms.

Cancel the common factor of $- 8 + 2 i \sqrt{3}$ and $40$ .

Factor $2$ out of $- 8$ .

$\frac{2 \cdot - 4 + 2 i \sqrt{3}}{40}$

Factor $2$ out of $2 i \sqrt{3}$ .

$\frac{2 \cdot - 4 + 2 (i \sqrt{3})}{40}$

Factor $2$ out of $2 \cdot - 4 + 2 (i \sqrt{3})$ .

$\frac{2 \cdot (- 4 + i \sqrt{3})}{40}$

Cancel the common factors.

Factor $2$ out of $40$ .

$\frac{2 \cdot (- 4 + i \sqrt{3})}{2 (20)}$

Cancel the common factor.

$\frac{2 \cdot (- 4 + i \sqrt{3})}{2 \cdot 20}$

Rewrite the expression.

$\frac{- 4 + i \sqrt{3}}{20}$

Rewrite $- 4$ as $- 1 (4)$ .

$\frac{- 1 (4) + i \sqrt{3}}{20}$

Factor $- 1$ out of $i \sqrt{3}$ .

$\frac{- 1 (4) - (- i \sqrt{3})}{20}$

Factor $- 1$ out of $- 1 (4) - (- i \sqrt{3})$ .

$\frac{- 1 (4 - i \sqrt{3})}{20}$

Move the negative in front of the fraction.

$- \frac{4 - i \sqrt{3}}{20}$