Solve 2x^2=18x-1

$$2 { x }^{ 2 } =18x-1$$

Solve for x

$$x = \frac{\sqrt{79} + 9}{2} \approx 8.944097209$$

$$x=\frac{9-\sqrt{79}}{2}\approx 0.055902791$$

Steps Using the Quadratic Formula

Subtract 18x from both sides.

$$2x^{2}-18x=-1$$

Add 1 to both sides.

$$2x^{2}-18x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -18 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -18.

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

Multiply -4 times 2.

$$x=\frac{-\left(-18\right)±\sqrt{324-8}}{2\times 2}$$

Add 324 to -8.

$$x=\frac{-\left(-18\right)±\sqrt{316}}{2\times 2}$$

Take the square root of 316.

$$x=\frac{-\left(-18\right)±2\sqrt{79}}{2\times 2}$$

The opposite of -18 is 18.

$$x=\frac{18±2\sqrt{79}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{18±2\sqrt{79}}{4}$$

Now solve the equation x=\frac{18±2\sqrt{79}}{4} when ± is plus. Add 18 to 2\sqrt{79}.

$$x=\frac{2\sqrt{79}+18}{4}$$

Divide 18+2\sqrt{79} by 4.

$$x=\frac{\sqrt{79}+9}{2}$$

Now solve the equation x=\frac{18±2\sqrt{79}}{4} when ± is minus. Subtract 2\sqrt{79} from 18.

$$x=\frac{18-2\sqrt{79}}{4}$$

Divide 18-2\sqrt{79} by 4.

$$x=\frac{9-\sqrt{79}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{79}+9}{2} x=\frac{9-\sqrt{79}}{2}$$

Steps for Completing the Square

Subtract 18x from both sides.

$$2x^{2}-18x=-1$$

Divide both sides by 2.

$$\frac{2x^{2}-18x}{2}=-\frac{1}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{18}{2}\right)x=-\frac{1}{2}$$

Divide -18 by 2.

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

Divide -9, the coefficient of the x term, by 2 to get -\frac{9}{2}. Then add the square of -\frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-9x+\frac{81}{4}=-\frac{1}{2}+\frac{81}{4}$$

Add -\frac{1}{2} to \frac{81}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-9x+\frac{81}{4}=\frac{79}{4}$$

Factor x^{2}-9x+\frac{81}{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}.

$$\left(x-\frac{9}{2}\right)^{2}=\frac{79}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{2}\right)^{2}}=\sqrt{\frac{79}{4}}$$

Simplify.

$$x-\frac{9}{2}=\frac{\sqrt{79}}{2} x-\frac{9}{2}=-\frac{\sqrt{79}}{2}$$

Add \frac{9}{2} to both sides of the equation.

$$x=\frac{\sqrt{79}+9}{2} x=\frac{9-\sqrt{79}}{2}$$