Solve (2)x^2+3=3sqrt{2}x

$$( 2 ) x ^ { 2 } + 3 = 3 \sqrt { 2 } x$$

Solve for x (complex solution)

$$x=\frac{3\sqrt{2}+\sqrt{6}i}{4}\approx 1.060660172+0.612372436i$$

$$x=\frac{-\sqrt{6}i+3\sqrt{2}}{4}\approx 1.060660172-0.612372436i$$

Steps Using the Quadratic Formula

Subtract 3\sqrt{2}x from both sides.

$$2x^{2}+3-3\sqrt{2}x=0$$

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.

$$2x^{2}+\left(-3\sqrt{2}\right)x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -3\sqrt{2} for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -3\sqrt{2}.

$$x=\frac{-\left(-3\sqrt{2}\right)±\sqrt{18-4\times 2\times 3}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-3\sqrt{2}\right)±\sqrt{18-8\times 3}}{2\times 2}$$

Multiply -8 times 3.

$$x=\frac{-\left(-3\sqrt{2}\right)±\sqrt{18-24}}{2\times 2}$$

Add 18 to -24.

$$x=\frac{-\left(-3\sqrt{2}\right)±\sqrt{-6}}{2\times 2}$$

Take the square root of -6.

$$x=\frac{-\left(-3\sqrt{2}\right)±\sqrt{6}i}{2\times 2}$$

The opposite of -3\sqrt{2} is 3\sqrt{2}.

$$x=\frac{3\sqrt{2}±\sqrt{6}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{3\sqrt{2}±\sqrt{6}i}{4}$$

Now solve the equation x=\frac{3\sqrt{2}±\sqrt{6}i}{4} when ± is plus. Add 3\sqrt{2} to i\sqrt{6}.

$$x=\frac{3\sqrt{2}+\sqrt{6}i}{4}$$

Now solve the equation x=\frac{3\sqrt{2}±\sqrt{6}i}{4} when ± is minus. Subtract i\sqrt{6} from 3\sqrt{2}.

$$x=\frac{-\sqrt{6}i+3\sqrt{2}}{4}$$

The equation is now solved.

$$x=\frac{3\sqrt{2}+\sqrt{6}i}{4} x=\frac{-\sqrt{6}i+3\sqrt{2}}{4}$$

Steps for Completing the Square

Subtract 3\sqrt{2}x from both sides.

$$2x^{2}+3-3\sqrt{2}x=0$$

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

$$2x^{2}-3\sqrt{2}x=-3$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide -\frac{3\sqrt{2}}{2}, the coefficient of the x term, by 2 to get -\frac{3\sqrt{2}}{4}. Then add the square of -\frac{3\sqrt{2}}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{3\sqrt{2}}{4}.

$$x^{2}+\left(-\frac{3\sqrt{2}}{2}\right)x+\frac{9}{8}=-\frac{3}{2}+\frac{9}{8}$$

Add -\frac{3}{2} to \frac{9}{8} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\left(-\frac{3\sqrt{2}}{2}\right)x+\frac{9}{8}=-\frac{3}{8}$$

Factor x^{2}+\left(-\frac{3\sqrt{2}}{2}\right)x+\frac{9}{8}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3\sqrt{2}}{4}\right)^{2}=-\frac{3}{8}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3\sqrt{2}}{4}\right)^{2}}=\sqrt{-\frac{3}{8}}$$

Simplify.

$$x-\frac{3\sqrt{2}}{4}=\frac{\sqrt{6}i}{4} x-\frac{3\sqrt{2}}{4}=-\frac{\sqrt{6}i}{4}$$

Add \frac{3\sqrt{2}}{4} to both sides of the equation.

$$x=\frac{3\sqrt{2}+\sqrt{6}i}{4} x=\frac{-\sqrt{6}i+3\sqrt{2}}{4}$$