Solve 3x-8-14/x=2/x

$$3x-8- \frac{ 14 }{ x } = \frac{ 2 }{ x }$$

Solve for x

$$x = -\frac{4}{3} = -1\frac{1}{3} \approx -1.333333333$$

$$x=4$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$3xx+x\left(-8\right)-14=2$$

Multiply x and x to get x^{2}.

$$3x^{2}+x\left(-8\right)-14=2$$

Subtract 2 from both sides.

$$3x^{2}+x\left(-8\right)-14-2=0$$

Subtract 2 from -14 to get -16.

$$3x^{2}+x\left(-8\right)-16=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$3x^{2}-8x-16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -8 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 3\left(-16\right)}}{2\times 3}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 3\left(-16\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-8\right)±\sqrt{64-12\left(-16\right)}}{2\times 3}$$

Multiply -12 times -16.

$$x=\frac{-\left(-8\right)±\sqrt{64+192}}{2\times 3}$$

Add 64 to 192.

$$x=\frac{-\left(-8\right)±\sqrt{256}}{2\times 3}$$

Take the square root of 256.

$$x=\frac{-\left(-8\right)±16}{2\times 3}$$

The opposite of -8 is 8.

$$x=\frac{8±16}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{8±16}{6}$$

Now solve the equation x=\frac{8±16}{6} when ± is plus. Add 8 to 16.

$$x=\frac{24}{6}$$

Divide 24 by 6.

$$x=4$$

Now solve the equation x=\frac{8±16}{6} when ± is minus. Subtract 16 from 8.

$$x=-\frac{8}{6}$$

Reduce the fraction \frac{-8}{6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{4}{3}$$

The equation is now solved.

$$x=4 x=-\frac{4}{3}$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$3xx+x\left(-8\right)-14=2$$

Multiply x and x to get x^{2}.

$$3x^{2}+x\left(-8\right)-14=2$$

Add 14 to both sides.

$$3x^{2}+x\left(-8\right)=2+14$$

Add 2 and 14 to get 16.

$$3x^{2}+x\left(-8\right)=16$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$3x^{2}-8x=16$$

Divide both sides by 3.

$$\frac{3x^{2}-8x}{3}=\frac{16}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{8}{3}x=\frac{16}{3}$$

Divide -\frac{8}{3}, the coefficient of the x term, by 2 to get -\frac{4}{3}. Then add the square of -\frac{4}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{8}{3}x+\left(-\frac{4}{3}\right)^{2}=\frac{16}{3}+\left(-\frac{4}{3}\right)^{2}$$

Square -\frac{4}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{8}{3}x+\frac{16}{9}=\frac{16}{3}+\frac{16}{9}$$

Add \frac{16}{3} to \frac{16}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{8}{3}x+\frac{16}{9}=\frac{64}{9}$$

Factor x^{2}-\frac{8}{3}x+\frac{16}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{4}{3}\right)^{2}=\frac{64}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{4}{3}\right)^{2}}=\sqrt{\frac{64}{9}}$$

Simplify.

$$x-\frac{4}{3}=\frac{8}{3} x-\frac{4}{3}=-\frac{8}{3}$$

Add \frac{4}{3} to both sides of the equation.

$$x=4 x=-\frac{4}{3}$$