Solve -2x+13=6/x

$$- 2 x + 13 = \frac { 6 } { x }$$

Solve for x

$$x=\frac{1}{2}=0.5$$

$$x=6$$

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.

$$-2xx+x\times 13=6$$

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

$$-2x^{2}+x\times 13=6$$

Subtract 6 from both sides.

$$-2x^{2}+x\times 13-6=0$$

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.

$$-2x^{2}+13x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 13 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-13±\sqrt{13^{2}-4\left(-2\right)\left(-6\right)}}{2\left(-2\right)}$$

Square 13.

$$x=\frac{-13±\sqrt{169-4\left(-2\right)\left(-6\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-13±\sqrt{169+8\left(-6\right)}}{2\left(-2\right)}$$

Multiply 8 times -6.

$$x=\frac{-13±\sqrt{169-48}}{2\left(-2\right)}$$

Add 169 to -48.

$$x=\frac{-13±\sqrt{121}}{2\left(-2\right)}$$

Take the square root of 121.

$$x=\frac{-13±11}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-13±11}{-4}$$

Now solve the equation x=\frac{-13±11}{-4} when ± is plus. Add -13 to 11.

$$x=-\frac{2}{-4}$$

Reduce the fraction \frac{-2}{-4} to lowest terms by extracting and canceling out 2.

$$x=\frac{1}{2}$$

Now solve the equation x=\frac{-13±11}{-4} when ± is minus. Subtract 11 from -13.

$$x=-\frac{24}{-4}$$

Divide -24 by -4.

$$x=6$$

The equation is now solved.

$$x=\frac{1}{2} x=6$$

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.

$$-2xx+x\times 13=6$$

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

$$-2x^{2}+x\times 13=6$$

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.

$$-2x^{2}+13x=6$$

Divide both sides by -2.

$$\frac{-2x^{2}+13x}{-2}=\frac{6}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{13}{-2}x=\frac{6}{-2}$$

Divide 13 by -2.

$$x^{2}-\frac{13}{2}x=\frac{6}{-2}$$

Divide 6 by -2.

$$x^{2}-\frac{13}{2}x=-3$$

Divide -\frac{13}{2}, the coefficient of the x term, by 2 to get -\frac{13}{4}. Then add the square of -\frac{13}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{13}{2}x+\left(-\frac{13}{4}\right)^{2}=-3+\left(-\frac{13}{4}\right)^{2}$$

Square -\frac{13}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{13}{2}x+\frac{169}{16}=-3+\frac{169}{16}$$

Add -3 to \frac{169}{16}.

$$x^{2}-\frac{13}{2}x+\frac{169}{16}=\frac{121}{16}$$

Factor x^{2}-\frac{13}{2}x+\frac{169}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{13}{4}\right)^{2}=\frac{121}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{13}{4}\right)^{2}}=\sqrt{\frac{121}{16}}$$

Simplify.

$$x-\frac{13}{4}=\frac{11}{4} x-\frac{13}{4}=-\frac{11}{4}$$

Add \frac{13}{4} to both sides of the equation.

$$x=6 x=\frac{1}{2}$$