Solve 3x^2=4x-4

$$3 x ^ { 2 } = 4 x - 4$$

Solve for x (complex solution)

$$x=\frac{2+2\sqrt{2}i}{3}\approx 0.666666667+0.942809042i$$

$$x=\frac{-2\sqrt{2}i+2}{3}\approx 0.666666667-0.942809042i$$

Steps Using the Quadratic Formula

Subtract 4x from both sides.

$$3x^{2}-4x=-4$$

Add 4 to both sides.

$$3x^{2}-4x+4=0$$

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

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

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 3\times 4}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-4\right)±\sqrt{16-12\times 4}}{2\times 3}$$

Multiply -12 times 4.

$$x=\frac{-\left(-4\right)±\sqrt{16-48}}{2\times 3}$$

Add 16 to -48.

$$x=\frac{-\left(-4\right)±\sqrt{-32}}{2\times 3}$$

Take the square root of -32.

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

The opposite of -4 is 4.

$$x=\frac{4±4\sqrt{2}i}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{4±4\sqrt{2}i}{6}$$

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

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

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

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

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

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

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

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

The equation is now solved.

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

Steps for Completing the Square

Subtract 4x from both sides.

$$3x^{2}-4x=-4$$

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

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

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

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

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

Add -\frac{4}{3} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=-\frac{8}{9}$$

Factor x^{2}-\frac{4}{3}x+\frac{4}{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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

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

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