Solve 2x^2=6x+18

$$2 { x }^{ 2 } =6x+18$$

Solve for x

$$x = \frac{3 \sqrt{5} + 3}{2} \approx 4.854101966$$

$$x=\frac{3-3\sqrt{5}}{2}\approx -1.854101966$$

Steps Using the Quadratic Formula

Subtract 6x from both sides.

$$2x^{2}-6x=18$$

Subtract 18 from both sides.

$$2x^{2}-6x-18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -6 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 2\left(-18\right)}}{2\times 2}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 2\left(-18\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-6\right)±\sqrt{36-8\left(-18\right)}}{2\times 2}$$

Multiply -8 times -18.

$$x=\frac{-\left(-6\right)±\sqrt{36+144}}{2\times 2}$$

Add 36 to 144.

$$x=\frac{-\left(-6\right)±\sqrt{180}}{2\times 2}$$

Take the square root of 180.

$$x=\frac{-\left(-6\right)±6\sqrt{5}}{2\times 2}$$

The opposite of -6 is 6.

$$x=\frac{6±6\sqrt{5}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{6±6\sqrt{5}}{4}$$

Now solve the equation x=\frac{6±6\sqrt{5}}{4} when ± is plus. Add 6 to 6\sqrt{5}.

$$x=\frac{6\sqrt{5}+6}{4}$$

Divide 6+6\sqrt{5} by 4.

$$x=\frac{3\sqrt{5}+3}{2}$$

Now solve the equation x=\frac{6±6\sqrt{5}}{4} when ± is minus. Subtract 6\sqrt{5} from 6.

$$x=\frac{6-6\sqrt{5}}{4}$$

Divide 6-6\sqrt{5} by 4.

$$x=\frac{3-3\sqrt{5}}{2}$$

The equation is now solved.

$$x=\frac{3\sqrt{5}+3}{2} x=\frac{3-3\sqrt{5}}{2}$$

Steps for Completing the Square

Subtract 6x from both sides.

$$2x^{2}-6x=18$$

Divide both sides by 2.

$$\frac{2x^{2}-6x}{2}=\frac{18}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{6}{2}\right)x=\frac{18}{2}$$

Divide -6 by 2.

$$x^{2}-3x=\frac{18}{2}$$

Divide 18 by 2.

$$x^{2}-3x=9$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=9+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=9+\frac{9}{4}$$

Add 9 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{45}{4}$$

Factor x^{2}-3x+\frac{9}{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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{45}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{45}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{3\sqrt{5}}{2} x-\frac{3}{2}=-\frac{3\sqrt{5}}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=\frac{3\sqrt{5}+3}{2} x=\frac{3-3\sqrt{5}}{2}$$