Solve x=3x^2-2

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

Solve for x

$$x=-\frac{2}{3}\approx -0.666666667$$

$$x=1$$

Steps Using Factoring By Grouping

Subtract 3x^{2} from both sides.

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

Add 2 to both sides.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx+2. To find a and b, set up a system to be solved.

$$a+b=1 ab=-3\times 2=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 1.

$$a=3 b=-2$$

Rewrite -3x^{2}+x+2 as \left(-3x^{2}+3x\right)+\left(-2x+2\right).

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

Factor out 3x in the first and 2 in the second group.

$$3x\left(-x+1\right)+2\left(-x+1\right)$$

Factor out common term -x+1 by using distributive property.

$$\left(-x+1\right)\left(3x+2\right)$$

To find equation solutions, solve -x+1=0 and 3x+2=0.

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

Steps Using the Quadratic Formula

Subtract 3x^{2} from both sides.

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

Add 2 to both sides.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 1 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\left(-3\right)\times 2}}{2\left(-3\right)}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\left(-3\right)\times 2}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-1±\sqrt{1+12\times 2}}{2\left(-3\right)}$$

Multiply 12 times 2.

$$x=\frac{-1±\sqrt{1+24}}{2\left(-3\right)}$$

Add 1 to 24.

$$x=\frac{-1±\sqrt{25}}{2\left(-3\right)}$$

Take the square root of 25.

$$x=\frac{-1±5}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-1±5}{-6}$$

Now solve the equation x=\frac{-1±5}{-6} when ± is plus. Add -1 to 5.

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

Reduce the fraction \frac{4}{-6} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-1±5}{-6} when ± is minus. Subtract 5 from -1.

$$x=-\frac{6}{-6}$$

Divide -6 by -6.

$$x=1$$

The equation is now solved.

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

Steps for Completing the Square

Subtract 3x^{2} from both sides.

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

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.

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

Divide both sides by -3.

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

Dividing by -3 undoes the multiplication by -3.

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

Divide 1 by -3.

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

Divide -2 by -3.

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

Divide -\frac{1}{3}, the coefficient of the x term, by 2 to get -\frac{1}{6}. Then add the square of -\frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=\frac{2}{3}+\left(-\frac{1}{6}\right)^{2}$$

Square -\frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{2}{3}+\frac{1}{36}$$

Add \frac{2}{3} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{25}{36}$$

Factor x^{2}-\frac{1}{3}x+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{6}\right)^{2}=\frac{25}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{\frac{25}{36}}$$

Simplify.

$$x-\frac{1}{6}=\frac{5}{6} x-\frac{1}{6}=-\frac{5}{6}$$

Add \frac{1}{6} to both sides of the equation.

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