Solve (x+5divx+2+x-5)+(x-5divx+2)=2

$$(x+5 \div x+2+x-5)+(x-5 \div x+2)=2$$

Solve for x

$$x=1$$

Steps Using Factoring

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

$$xx+5+x\times 2+xx+x\left(-5\right)+xx-5+x\times 2=2x$$

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

$$x^{2}+5+x\times 2+xx+x\left(-5\right)+xx-5+x\times 2=2x$$

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

$$x^{2}+5+x\times 2+x^{2}+x\left(-5\right)+xx-5+x\times 2=2x$$

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

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

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

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

Combine x\times 2 and x\left(-5\right) to get -3x.

$$2x^{2}+5-3x+x^{2}-5+x\times 2=2x$$

Combine 2x^{2} and x^{2} to get 3x^{2}.

$$3x^{2}+5-3x-5+x\times 2=2x$$

Subtract 5 from 5 to get 0.

$$3x^{2}-3x+x\times 2=2x$$

Combine -3x and x\times 2 to get -x.

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

Subtract 2x from both sides.

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

Combine -x and -2x to get -3x.

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

Factor out x.

$$x\left(3x-3\right)=0$$

To find equation solutions, solve x=0 and 3x-3=0.

$$x=0 x=1$$

Variable x cannot be equal to 0.

$$x=1$$

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.

$$xx+5+x\times 2+xx+x\left(-5\right)+xx-5+x\times 2=2x$$

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

$$x^{2}+5+x\times 2+xx+x\left(-5\right)+xx-5+x\times 2=2x$$

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

$$x^{2}+5+x\times 2+x^{2}+x\left(-5\right)+xx-5+x\times 2=2x$$

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

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

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

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

Combine x\times 2 and x\left(-5\right) to get -3x.

$$2x^{2}+5-3x+x^{2}-5+x\times 2=2x$$

Combine 2x^{2} and x^{2} to get 3x^{2}.

$$3x^{2}+5-3x-5+x\times 2=2x$$

Subtract 5 from 5 to get 0.

$$3x^{2}-3x+x\times 2=2x$$

Combine -3x and x\times 2 to get -x.

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

Subtract 2x from both sides.

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

Combine -x and -2x to get -3x.

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

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

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

Take the square root of \left(-3\right)^{2}.

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

The opposite of -3 is 3.

$$x=\frac{3±3}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{3±3}{6}$$

Now solve the equation x=\frac{3±3}{6} when ± is plus. Add 3 to 3.

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

Divide 6 by 6.

$$x=1$$

Now solve the equation x=\frac{3±3}{6} when ± is minus. Subtract 3 from 3.

$$x=\frac{0}{6}$$

Divide 0 by 6.

$$x=0$$

The equation is now solved.

$$x=1 x=0$$

Variable x cannot be equal to 0.

$$x=1$$

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.

$$xx+5+x\times 2+xx+x\left(-5\right)+xx-5+x\times 2=2x$$

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

$$x^{2}+5+x\times 2+xx+x\left(-5\right)+xx-5+x\times 2=2x$$

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

$$x^{2}+5+x\times 2+x^{2}+x\left(-5\right)+xx-5+x\times 2=2x$$

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

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

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

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

Combine x\times 2 and x\left(-5\right) to get -3x.

$$2x^{2}+5-3x+x^{2}-5+x\times 2=2x$$

Combine 2x^{2} and x^{2} to get 3x^{2}.

$$3x^{2}+5-3x-5+x\times 2=2x$$

Subtract 5 from 5 to get 0.

$$3x^{2}-3x+x\times 2=2x$$

Combine -3x and x\times 2 to get -x.

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

Subtract 2x from both sides.

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

Combine -x and -2x to get -3x.

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

Divide both sides by 3.

$$\frac{3x^{2}-3x}{3}=\frac{0}{3}$$

Dividing by 3 undoes the multiplication by 3.

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

Divide -3 by 3.

$$x^{2}-x=\frac{0}{3}$$

Divide 0 by 3.

$$x^{2}-x=0$$

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Factor x^{2}-x+\frac{1}{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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{1}{2} to both sides of the equation.

$$x=1 x=0$$

Variable x cannot be equal to 0.

$$x=1$$