Solve 2=3x^2-6x+5

$$2 = 3 x ^ { 2 } - 6 x + 5$$

Solve for x

$$x=1$$

Steps Using Factoring By Grouping

Swap sides so that all variable terms are on the left hand side.

$$3x^{2}-6x+5=2$$

Subtract 2 from both sides.

$$3x^{2}-6x+5-2=0$$

Subtract 2 from 5 to get 3.

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

Divide both sides by 3.

$$x^{2}-2x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

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

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-1 b=-1$$

Rewrite x^{2}-2x+1 as \left(x^{2}-x\right)+\left(-x+1\right).

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

Factor out x in the first and -1 in the second group.

$$x\left(x-1\right)-\left(x-1\right)$$

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

$$\left(x-1\right)\left(x-1\right)$$

Rewrite as a binomial square.

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

To find equation solution, solve x-1=0.

$$x=1$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$3x^{2}-6x+5=2$$

Subtract 2 from both sides.

$$3x^{2}-6x+5-2=0$$

Subtract 2 from 5 to get 3.

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

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

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

Square -6.

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

Multiply -4 times 3.

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

Multiply -12 times 3.

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

Add 36 to -36.

$$x=\frac{-\left(-6\right)±\sqrt{0}}{2\times 3}$$

Take the square root of 0.

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

The opposite of -6 is 6.

$$x=\frac{6}{2\times 3}$$

Multiply 2 times 3.

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

Divide 6 by 6.

$$x=1$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$3x^{2}-6x+5=2$$

Subtract 5 from both sides.

$$3x^{2}-6x=2-5$$

Subtract 5 from 2 to get -3.

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

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

Divide -6 by 3.

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

Divide -3 by 3.

$$x^{2}-2x=-1$$

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.

$$x^{2}-2x+1=-1+1$$

Add -1 to 1.

$$x^{2}-2x+1=0$$

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}.

$$\left(x-1\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-1=0 x-1=0$$

Add 1 to both sides of the equation.

$$x=1 x=1$$

The equation is now solved. Solutions are the same.

$$x=1$$