Solve 3x^2+3=-2x

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

Solve for x (complex solution)

$$x=\frac{-1+2\sqrt{2}i}{3}\approx -0.333333333+0.942809042i$$

$$x=\frac{-2\sqrt{2}i-1}{3}\approx -0.333333333-0.942809042i$$

Steps Using the Quadratic Formula

Add 2x to both sides.

$$3x^{2}+3+2x=0$$

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.

$$3x^{2}+2x+3=0$$

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

$$x=\frac{-2±\sqrt{2^{2}-4\times 3\times 3}}{2\times 3}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times 3\times 3}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-2±\sqrt{4-12\times 3}}{2\times 3}$$

Multiply -12 times 3.

$$x=\frac{-2±\sqrt{4-36}}{2\times 3}$$

Add 4 to -36.

$$x=\frac{-2±\sqrt{-32}}{2\times 3}$$

Take the square root of -32.

$$x=\frac{-2±4\sqrt{2}i}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-2±4\sqrt{2}i}{6}$$

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

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

Divide -2+4i\sqrt{2} by 6.

$$x=\frac{-1+2\sqrt{2}i}{3}$$

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

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

Divide -2-4i\sqrt{2} by 6.

$$x=\frac{-2\sqrt{2}i-1}{3}$$

The equation is now solved.

$$x=\frac{-1+2\sqrt{2}i}{3} x=\frac{-2\sqrt{2}i-1}{3}$$

Steps for Completing the Square

Add 2x to both sides.

$$3x^{2}+3+2x=0$$

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

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

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

Divide -3 by 3.

$$x^{2}+\frac{2}{3}x=-1$$

Divide \frac{2}{3}, the coefficient of the x term, by 2 to get \frac{1}{3}. Then add the square of \frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{2}{3}x+\left(\frac{1}{3}\right)^{2}=-1+\left(\frac{1}{3}\right)^{2}$$

Square \frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=-1+\frac{1}{9}$$

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

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=-\frac{8}{9}$$

Factor x^{2}+\frac{2}{3}x+\frac{1}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{3}\right)^{2}=-\frac{8}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{-\frac{8}{9}}$$

Simplify.

$$x+\frac{1}{3}=\frac{2\sqrt{2}i}{3} x+\frac{1}{3}=-\frac{2\sqrt{2}i}{3}$$

Subtract \frac{1}{3} from both sides of the equation.

$$x=\frac{-1+2\sqrt{2}i}{3} x=\frac{-2\sqrt{2}i-1}{3}$$