Solve 2x^2=22x-11

$$2 x ^ { 2 } = 22 x - 11$$

Solve for x

$$x = \frac{3 \sqrt{11} + 11}{2} \approx 10.474937186$$

$$x=\frac{11-3\sqrt{11}}{2}\approx 0.525062814$$

Steps Using the Quadratic Formula

Subtract 22x from both sides.

$$2x^{2}-22x=-11$$

Add 11 to both sides.

$$2x^{2}-22x+11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -22 for b, and 11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-22\right)±\sqrt{\left(-22\right)^{2}-4\times 2\times 11}}{2\times 2}$$

Square -22.

$$x=\frac{-\left(-22\right)±\sqrt{484-4\times 2\times 11}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-22\right)±\sqrt{484-8\times 11}}{2\times 2}$$

Multiply -8 times 11.

$$x=\frac{-\left(-22\right)±\sqrt{484-88}}{2\times 2}$$

Add 484 to -88.

$$x=\frac{-\left(-22\right)±\sqrt{396}}{2\times 2}$$

Take the square root of 396.

$$x=\frac{-\left(-22\right)±6\sqrt{11}}{2\times 2}$$

The opposite of -22 is 22.

$$x=\frac{22±6\sqrt{11}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{22±6\sqrt{11}}{4}$$

Now solve the equation x=\frac{22±6\sqrt{11}}{4} when ± is plus. Add 22 to 6\sqrt{11}.

$$x=\frac{6\sqrt{11}+22}{4}$$

Divide 22+6\sqrt{11} by 4.

$$x=\frac{3\sqrt{11}+11}{2}$$

Now solve the equation x=\frac{22±6\sqrt{11}}{4} when ± is minus. Subtract 6\sqrt{11} from 22.

$$x=\frac{22-6\sqrt{11}}{4}$$

Divide 22-6\sqrt{11} by 4.

$$x=\frac{11-3\sqrt{11}}{2}$$

The equation is now solved.

$$x=\frac{3\sqrt{11}+11}{2} x=\frac{11-3\sqrt{11}}{2}$$

Steps for Completing the Square

Subtract 22x from both sides.

$$2x^{2}-22x=-11$$

Divide both sides by 2.

$$\frac{2x^{2}-22x}{2}=-\frac{11}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{22}{2}\right)x=-\frac{11}{2}$$

Divide -22 by 2.

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

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

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

Square -\frac{11}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-11x+\frac{121}{4}=-\frac{11}{2}+\frac{121}{4}$$

Add -\frac{11}{2} to \frac{121}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-11x+\frac{121}{4}=\frac{99}{4}$$

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

$$\left(x-\frac{11}{2}\right)^{2}=\frac{99}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{2}\right)^{2}}=\sqrt{\frac{99}{4}}$$

Simplify.

$$x-\frac{11}{2}=\frac{3\sqrt{11}}{2} x-\frac{11}{2}=-\frac{3\sqrt{11}}{2}$$

Add \frac{11}{2} to both sides of the equation.

$$x=\frac{3\sqrt{11}+11}{2} x=\frac{11-3\sqrt{11}}{2}$$