Solve a^2+9/4=(2a-3/2)^2

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

Solve for a

$$a=2$$

$$a=0$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2a-\frac{3}{2}\right)^{2}.

$$a^{2}+\frac{9}{4}=4a^{2}-6a+\frac{9}{4}$$

Subtract 4a^{2} from both sides.

$$a^{2}+\frac{9}{4}-4a^{2}=-6a+\frac{9}{4}$$

Combine a^{2} and -4a^{2} to get -3a^{2}.

$$-3a^{2}+\frac{9}{4}=-6a+\frac{9}{4}$$

Add 6a to both sides.

$$-3a^{2}+\frac{9}{4}+6a=\frac{9}{4}$$

Subtract \frac{9}{4} from both sides.

$$-3a^{2}+\frac{9}{4}+6a-\frac{9}{4}=0$$

Subtract \frac{9}{4} from \frac{9}{4} to get 0.

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

Factor out a.

$$a\left(-3a+6\right)=0$$

To find equation solutions, solve a=0 and -3a+6=0.

$$a=0 a=2$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2a-\frac{3}{2}\right)^{2}.

$$a^{2}+\frac{9}{4}=4a^{2}-6a+\frac{9}{4}$$

Subtract 4a^{2} from both sides.

$$a^{2}+\frac{9}{4}-4a^{2}=-6a+\frac{9}{4}$$

Combine a^{2} and -4a^{2} to get -3a^{2}.

$$-3a^{2}+\frac{9}{4}=-6a+\frac{9}{4}$$

Add 6a to both sides.

$$-3a^{2}+\frac{9}{4}+6a=\frac{9}{4}$$

Subtract \frac{9}{4} from both sides.

$$-3a^{2}+\frac{9}{4}+6a-\frac{9}{4}=0$$

Subtract \frac{9}{4} from \frac{9}{4} to get 0.

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

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

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

Take the square root of 6^{2}.

$$a=\frac{-6±6}{2\left(-3\right)}$$

Multiply 2 times -3.

$$a=\frac{-6±6}{-6}$$

Now solve the equation a=\frac{-6±6}{-6} when ± is plus. Add -6 to 6.

$$a=\frac{0}{-6}$$

Divide 0 by -6.

$$a=0$$

Now solve the equation a=\frac{-6±6}{-6} when ± is minus. Subtract 6 from -6.

$$a=-\frac{12}{-6}$$

Divide -12 by -6.

$$a=2$$

The equation is now solved.

$$a=0 a=2$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2a-\frac{3}{2}\right)^{2}.

$$a^{2}+\frac{9}{4}=4a^{2}-6a+\frac{9}{4}$$

Subtract 4a^{2} from both sides.

$$a^{2}+\frac{9}{4}-4a^{2}=-6a+\frac{9}{4}$$

Combine a^{2} and -4a^{2} to get -3a^{2}.

$$-3a^{2}+\frac{9}{4}=-6a+\frac{9}{4}$$

Add 6a to both sides.

$$-3a^{2}+\frac{9}{4}+6a=\frac{9}{4}$$

Subtract \frac{9}{4} from both sides.

$$-3a^{2}+6a=\frac{9}{4}-\frac{9}{4}$$

Subtract \frac{9}{4} from \frac{9}{4} to get 0.

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

Divide both sides by -3.

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

Dividing by -3 undoes the multiplication by -3.

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

Divide 6 by -3.

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

Divide 0 by -3.

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

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.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add 1 to both sides of the equation.

$$a=2 a=0$$