Solve -6/x=-x+5

$$- \frac{ 6 }{ x } = -x+5$$

Solve for x

$$x=-1$$

$$x=6$$

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.

$$-6=-xx+x\times 5$$

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

$$-6=-x^{2}+x\times 5$$

Swap sides so that all variable terms are on the left hand side.

$$-x^{2}+x\times 5=-6$$

Add 6 to both sides.

$$-x^{2}+x\times 5+6=0$$

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.

$$-x^{2}+5x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 5 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-1\right)\times 6}}{2\left(-1\right)}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-1\right)\times 6}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-5±\sqrt{25+4\times 6}}{2\left(-1\right)}$$

Multiply 4 times 6.

$$x=\frac{-5±\sqrt{25+24}}{2\left(-1\right)}$$

Add 25 to 24.

$$x=\frac{-5±\sqrt{49}}{2\left(-1\right)}$$

Take the square root of 49.

$$x=\frac{-5±7}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-5±7}{-2}$$

Now solve the equation x=\frac{-5±7}{-2} when ± is plus. Add -5 to 7.

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

Divide 2 by -2.

$$x=-1$$

Now solve the equation x=\frac{-5±7}{-2} when ± is minus. Subtract 7 from -5.

$$x=-\frac{12}{-2}$$

Divide -12 by -2.

$$x=6$$

The equation is now solved.

$$x=-1 x=6$$

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.

$$-6=-xx+x\times 5$$

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

$$-6=-x^{2}+x\times 5$$

Swap sides so that all variable terms are on the left hand side.

$$-x^{2}+x\times 5=-6$$

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.

$$-x^{2}+5x=-6$$

Divide both sides by -1.

$$\frac{-x^{2}+5x}{-1}=-\frac{6}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{5}{-1}x=-\frac{6}{-1}$$

Divide 5 by -1.

$$x^{2}-5x=-\frac{6}{-1}$$

Divide -6 by -1.

$$x^{2}-5x=6$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=6+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=6+\frac{25}{4}$$

Add 6 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{49}{4}$$

Factor x^{2}-5x+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{7}{2} x-\frac{5}{2}=-\frac{7}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=6 x=-1$$