Solve 3+2a=a^2

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

Solve for a

$$a=-1$$

$$a=3$$

Steps Using Factoring By Grouping

Subtract a^{2} from both sides.

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

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

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -a^{2}+aa+ba+3. To find a and b, set up a system to be solved.

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

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=3 b=-1$$

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

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

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

$$-a\left(a-3\right)-\left(a-3\right)$$

Factor out common term a-3 by using distributive property.

$$\left(a-3\right)\left(-a-1\right)$$

To find equation solutions, solve a-3=0 and -a-1=0.

$$a=3 a=-1$$

Steps Using the Quadratic Formula

Subtract a^{2} from both sides.

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

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.

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

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

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

Square 2.

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

Multiply -4 times -1.

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

Multiply 4 times 3.

$$a=\frac{-2±\sqrt{4+12}}{2\left(-1\right)}$$

Add 4 to 12.

$$a=\frac{-2±\sqrt{16}}{2\left(-1\right)}$$

Take the square root of 16.

$$a=\frac{-2±4}{2\left(-1\right)}$$

Multiply 2 times -1.

$$a=\frac{-2±4}{-2}$$

Now solve the equation a=\frac{-2±4}{-2} when ± is plus. Add -2 to 4.

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

Divide 2 by -2.

$$a=-1$$

Now solve the equation a=\frac{-2±4}{-2} when ± is minus. Subtract 4 from -2.

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

Divide -6 by -2.

$$a=3$$

The equation is now solved.

$$a=-1 a=3$$

Steps for Completing the Square

Subtract a^{2} from both sides.

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

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

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

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.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 2 by -1.

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

Divide -3 by -1.

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

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.

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

Add 3 to 1.

$$a^{2}-2a+1=4$$

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add 1 to both sides of the equation.

$$a=3 a=-1$$