Solve 6/1-4-3x/x^2-4=

$$\frac { 6 } { 1 - 4 } - \frac { 3 x } { x ^ { 2 } - 4 } =$$

Evaluate

$$\frac{8-3x-2x^{2}}{x^{2}-4}$$

Short Solution Steps

Subtract 4 from 1 to get -3.

$$\frac{6}{-3}-\frac{3x}{x^{2}-4}$$

Divide 6 by -3 to get -2.

$$-2-\frac{3x}{x^{2}-4}$$

Factor x^{2}-4.

$$-2-\frac{3x}{\left(x-2\right)\left(x+2\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -2 times \frac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}.

$$-\frac{2\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{3x}{\left(x-2\right)\left(x+2\right)}$$

Since -\frac{2\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)} and \frac{3x}{\left(x-2\right)\left(x+2\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-2\left(x-2\right)\left(x+2\right)-3x}{\left(x-2\right)\left(x+2\right)}$$

Do the multiplications in -2\left(x-2\right)\left(x+2\right)-3x.

$$\frac{-2x^{2}-4x+4x+8-3x}{\left(x-2\right)\left(x+2\right)}$$

Combine like terms in -2x^{2}-4x+4x+8-3x.

$$\frac{-2x^{2}-3x+8}{\left(x-2\right)\left(x+2\right)}$$

Expand \left(x-2\right)\left(x+2\right).

$$\frac{-2x^{2}-3x+8}{x^{2}-4}$$

Factor

$$-\frac{2\left(x-\frac{-\sqrt{73}-3}{4}\right)\left(x-\frac{\sqrt{73}-3}{4}\right)}{x^{2}-4}$$