Solve 3a-2=a^2

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

Solve for a

$$a=1$$

$$a=2$$

Steps Using Factoring By Grouping

Subtract a^{2} from both sides.

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

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.

$$a+b=3 ab=-\left(-2\right)=2$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. 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 -a^{2}+2a.

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

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

Subtract a^{2} from both sides.

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

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.

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

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}.

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

Square 3.

$$a=\frac{-3±\sqrt{9-4\left(-1\right)\left(-2\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$a=\frac{-3±\sqrt{9+4\left(-2\right)}}{2\left(-1\right)}$$

Multiply 4 times -2.

$$a=\frac{-3±\sqrt{9-8}}{2\left(-1\right)}$$

Add 9 to -8.

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

Take the square root of 1.

$$a=\frac{-3±1}{2\left(-1\right)}$$

Multiply 2 times -1.

$$a=\frac{-3±1}{-2}$$

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

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

Divide -2 by -2.

$$a=1$$

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

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

Divide -4 by -2.

$$a=2$$

The equation is now solved.

$$a=1 a=2$$

Steps for Completing the Square

Subtract a^{2} from both sides.

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

Add 2 to both sides. Anything plus zero gives itself.

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

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.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 3 by -1.

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

Divide 2 by -1.

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

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.

$$a=2 a=1$$