Solve 3x^2+2=6x

$$3 x ^ { 2 } + 2 = 6 x$$

Solve for x

$$x=\frac{\sqrt{3}}{3}+1\approx 1.577350269$$

$$x=-\frac{\sqrt{3}}{3}+1\approx 0.422649731$$

Steps Using the Quadratic Formula

Subtract 6x from both sides.

$$3x^{2}+2-6x=0$$

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.

$$3x^{2}-6x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -6 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 3\times 2}}{2\times 3}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 3\times 2}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-6\right)±\sqrt{36-12\times 2}}{2\times 3}$$

Multiply -12 times 2.

$$x=\frac{-\left(-6\right)±\sqrt{36-24}}{2\times 3}$$

Add 36 to -24.

$$x=\frac{-\left(-6\right)±\sqrt{12}}{2\times 3}$$

Take the square root of 12.

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

The opposite of -6 is 6.

$$x=\frac{6±2\sqrt{3}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{6±2\sqrt{3}}{6}$$

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

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

Divide 6+2\sqrt{3} by 6.

$$x=\frac{\sqrt{3}}{3}+1$$

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

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

Divide 6-2\sqrt{3} by 6.

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

The equation is now solved.

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

Steps for Completing the Square

Subtract 6x from both sides.

$$3x^{2}+2-6x=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$3x^{2}-6x=-2$$

Divide both sides by 3.

$$\frac{3x^{2}-6x}{3}=-\frac{2}{3}$$

Dividing by 3 undoes the multiplication by 3.

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

Divide -6 by 3.

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

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

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

Add -\frac{2}{3} to 1.

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

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

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add 1 to both sides of the equation.

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