Solve 9/4=a^2

$$\frac { 9 } { 4 } = a ^ { 2 }$$

Solve for a

$$a = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

$$a = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

Steps Using Difference of Squares

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

$$a^{2}=\frac{9}{4}$$

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

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

Multiply both sides by 4.

$$4a^{2}-9=0$$

Consider 4a^{2}-9. Rewrite 4a^{2}-9 as \left(2a\right)^{2}-3^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(2a-3\right)\left(2a+3\right)=0$$

To find equation solutions, solve 2a-3=0 and 2a+3=0.

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

Steps by Finding Square Root

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

$$a^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

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

Steps Using the Quadratic Formula

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

$$a^{2}=\frac{9}{4}$$

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

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -\frac{9}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{0±\sqrt{0^{2}-4\left(-\frac{9}{4}\right)}}{2}$$

Square 0.

$$a=\frac{0±\sqrt{-4\left(-\frac{9}{4}\right)}}{2}$$

Multiply -4 times -\frac{9}{4}.

$$a=\frac{0±\sqrt{9}}{2}$$

Take the square root of 9.

$$a=\frac{0±3}{2}$$

Now solve the equation a=\frac{0±3}{2} when ± is plus. Divide 3 by 2.

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

Now solve the equation a=\frac{0±3}{2} when ± is minus. Divide -3 by 2.

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

The equation is now solved.

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