Solve 1=3+a^2-3a

$$1 = 3 + a ^ { 2 } - 3 a$$

Solve for a

$$a=1$$

$$a=2$$

Steps Using Factoring

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

$$3+a^{2}-3a=1$$

Subtract 1 from both sides.

$$3+a^{2}-3a-1=0$$

Subtract 1 from 3 to get 2.

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

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

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

To solve the equation, factor a^{2}-3a+2 using formula a^{2}+\left(a+b\right)a+ab=\left(a+a\right)\left(a+b\right). To find a and b, set up a system to be solved.

$$a+b=-3 ab=2$$

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.

$$a=-2 b=-1$$

Rewrite factored expression \left(a+a\right)\left(a+b\right) using the obtained values.

$$\left(a-2\right)\left(a-1\right)$$

To find equation solutions, solve a-2=0 and a-1=0.

$$a=2 a=1$$

Steps Using Factoring By Grouping

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

$$3+a^{2}-3a=1$$

Subtract 1 from both sides.

$$3+a^{2}-3a-1=0$$

Subtract 1 from 3 to get 2.

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

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

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

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

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

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.

$$a=-2 b=-1$$

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

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

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

$$a\left(a-2\right)-\left(a-2\right)$$

Factor out common term a-2 by using distributive property.

$$\left(a-2\right)\left(a-1\right)$$

To find equation solutions, solve a-2=0 and a-1=0.

$$a=2 a=1$$

Steps Using the Quadratic Formula

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

$$3+a^{2}-3a=1$$

Subtract 1 from both sides.

$$3+a^{2}-3a-1=0$$

Subtract 1 from 3 to get 2.

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

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.

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

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

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

Square -3.

$$a=\frac{-\left(-3\right)±\sqrt{9-4\times 2}}{2}$$

Multiply -4 times 2.

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

Add 9 to -8.

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

Take the square root of 1.

$$a=\frac{-\left(-3\right)±1}{2}$$

The opposite of -3 is 3.

$$a=\frac{3±1}{2}$$

Now solve the equation a=\frac{3±1}{2} when ± is plus. Add 3 to 1.

$$a=\frac{4}{2}$$

Divide 4 by 2.

$$a=2$$

Now solve the equation a=\frac{3±1}{2} when ± is minus. Subtract 1 from 3.

$$a=\frac{2}{2}$$

Divide 2 by 2.

$$a=1$$

The equation is now solved.

$$a=2 a=1$$

Steps for Completing the Square

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

$$3+a^{2}-3a=1$$

Subtract 3 from both sides.

$$a^{2}-3a=1-3$$

Subtract 3 from 1 to get -2.

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

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

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

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

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

Add -2 to \frac{9}{4}.

$$a^{2}-3a+\frac{9}{4}=\frac{1}{4}$$

Factor a^{2}-3a+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a-\frac{3}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

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

Simplify.

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

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

$$a=2 a=1$$