Solve 2xleft(x-5right)+2=x

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

Solve for x

$$x = \frac{\sqrt{105} + 11}{4} \approx 5.311737691$$

$$x=\frac{11-\sqrt{105}}{4}\approx 0.188262309$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2x by x-5.

$$2x^{2}-10x+2=x$$

Subtract x from both sides.

$$2x^{2}-10x+2-x=0$$

Combine -10x and -x to get -11x.

$$2x^{2}-11x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -11 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 2\times 2}}{2\times 2}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 2\times 2}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-11\right)±\sqrt{121-8\times 2}}{2\times 2}$$

Multiply -8 times 2.

$$x=\frac{-\left(-11\right)±\sqrt{121-16}}{2\times 2}$$

Add 121 to -16.

$$x=\frac{-\left(-11\right)±\sqrt{105}}{2\times 2}$$

The opposite of -11 is 11.

$$x=\frac{11±\sqrt{105}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{11±\sqrt{105}}{4}$$

Now solve the equation x=\frac{11±\sqrt{105}}{4} when ± is plus. Add 11 to \sqrt{105}.

$$x=\frac{\sqrt{105}+11}{4}$$

Now solve the equation x=\frac{11±\sqrt{105}}{4} when ± is minus. Subtract \sqrt{105} from 11.

$$x=\frac{11-\sqrt{105}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{105}+11}{4} x=\frac{11-\sqrt{105}}{4}$$

Steps for Completing the Square

Use the distributive property to multiply 2x by x-5.

$$2x^{2}-10x+2=x$$

Subtract x from both sides.

$$2x^{2}-10x+2-x=0$$

Combine -10x and -x to get -11x.

$$2x^{2}-11x+2=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$2x^{2}-11x=-2$$

Divide both sides by 2.

$$\frac{2x^{2}-11x}{2}=-\frac{2}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{11}{2}x=-\frac{2}{2}$$

Divide -2 by 2.

$$x^{2}-\frac{11}{2}x=-1$$

Divide -\frac{11}{2}, the coefficient of the x term, by 2 to get -\frac{11}{4}. Then add the square of -\frac{11}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{2}x+\left(-\frac{11}{4}\right)^{2}=-1+\left(-\frac{11}{4}\right)^{2}$$

Square -\frac{11}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=-1+\frac{121}{16}$$

Add -1 to \frac{121}{16}.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=\frac{105}{16}$$

Factor x^{2}-\frac{11}{2}x+\frac{121}{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}.

$$\left(x-\frac{11}{4}\right)^{2}=\frac{105}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{4}\right)^{2}}=\sqrt{\frac{105}{16}}$$

Simplify.

$$x-\frac{11}{4}=\frac{\sqrt{105}}{4} x-\frac{11}{4}=-\frac{\sqrt{105}}{4}$$

Add \frac{11}{4} to both sides of the equation.

$$x=\frac{\sqrt{105}+11}{4} x=\frac{11-\sqrt{105}}{4}$$