Solve 9x^2-2=18x

$$9 x ^ { 2 } - 2 = 18 x$$

Solve for x

$$x=\frac{\sqrt{11}}{3}+1\approx 2.105541597$$

$$x=-\frac{\sqrt{11}}{3}+1\approx -0.105541597$$

Steps Using the Quadratic Formula

Subtract 18x from both sides.

$$9x^{2}-2-18x=0$$

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.

$$9x^{2}-18x-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -18 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 9\left(-2\right)}}{2\times 9}$$

Square -18.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\times 9\left(-2\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-18\right)±\sqrt{324-36\left(-2\right)}}{2\times 9}$$

Multiply -36 times -2.

$$x=\frac{-\left(-18\right)±\sqrt{324+72}}{2\times 9}$$

Add 324 to 72.

$$x=\frac{-\left(-18\right)±\sqrt{396}}{2\times 9}$$

Take the square root of 396.

$$x=\frac{-\left(-18\right)±6\sqrt{11}}{2\times 9}$$

The opposite of -18 is 18.

$$x=\frac{18±6\sqrt{11}}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{18±6\sqrt{11}}{18}$$

Now solve the equation x=\frac{18±6\sqrt{11}}{18} when ± is plus. Add 18 to 6\sqrt{11}.

$$x=\frac{6\sqrt{11}+18}{18}$$

Divide 18+6\sqrt{11} by 18.

$$x=\frac{\sqrt{11}}{3}+1$$

Now solve the equation x=\frac{18±6\sqrt{11}}{18} when ± is minus. Subtract 6\sqrt{11} from 18.

$$x=\frac{18-6\sqrt{11}}{18}$$

Divide 18-6\sqrt{11} by 18.

$$x=-\frac{\sqrt{11}}{3}+1$$

The equation is now solved.

$$x=\frac{\sqrt{11}}{3}+1 x=-\frac{\sqrt{11}}{3}+1$$

Steps for Completing the Square

Subtract 18x from both sides.

$$9x^{2}-2-18x=0$$

Add 2 to both sides. Anything plus zero gives itself.

$$9x^{2}-18x=2$$

Divide both sides by 9.

$$\frac{9x^{2}-18x}{9}=\frac{2}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\left(-\frac{18}{9}\right)x=\frac{2}{9}$$

Divide -18 by 9.

$$x^{2}-2x=\frac{2}{9}$$

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

$$x^{2}-2x+1=\frac{2}{9}+1$$

Add \frac{2}{9} to 1.

$$x^{2}-2x+1=\frac{11}{9}$$

Factor x^{2}-2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-1\right)^{2}=\frac{11}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{\frac{11}{9}}$$

Simplify.

$$x-1=\frac{\sqrt{11}}{3} x-1=-\frac{\sqrt{11}}{3}$$

Add 1 to both sides of the equation.

$$x=\frac{\sqrt{11}}{3}+1 x=-\frac{\sqrt{11}}{3}+1$$