Solve 3a^2=6a-3

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

Solve for a

$$a=1$$

Steps Using Factoring By Grouping

Subtract 6a from both sides.

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

Add 3 to both sides.

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

Divide both sides by 3.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as a^{2}+aa+ba+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 a^{2}-2a+1 as \left(a^{2}-a\right)+\left(-a+1\right).

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

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

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

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

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

Rewrite as a binomial square.

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

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

$$a=1$$

Steps Using the Quadratic Formula

Subtract 6a from both sides.

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

Add 3 to both sides.

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

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}.

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

Square -6.

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

Multiply -4 times 3.

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

Multiply -12 times 3.

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

Add 36 to -36.

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

Take the square root of 0.

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

The opposite of -6 is 6.

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

Multiply 2 times 3.

$$a=\frac{6}{6}$$

Divide 6 by 6.

$$a=1$$

Steps for Completing the Square

Subtract 6a from both sides.

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

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

Divide -6 by 3.

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

Divide -3 by 3.

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

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.

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

Add -1 to 1.

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

Factor a^{2}-2a+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$a-1=0 a-1=0$$

Add 1 to both sides of the equation.

$$a=1 a=1$$

The equation is now solved. Solutions are the same.

$$a=1$$