Solve (a+3)^2=(a+3)*(a-3)

$$( a + 3 ) ^ { 2 } = ( a + 3 ) \cdot ( a - 3 )$$

Solve for a

$$a=-3$$

Steps for Solving Linear Equation

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

$$a^{2}+6a+9=\left(a+3\right)\left(a-3\right)$$

Consider \left(a+3\right)\left(a-3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$a^{2}+6a+9=a^{2}-9$$

Subtract a^{2} from both sides.

$$a^{2}+6a+9-a^{2}=-9$$

Combine a^{2} and -a^{2} to get 0.

$$6a+9=-9$$

Subtract 9 from both sides.

$$6a=-9-9$$

Subtract 9 from -9 to get -18.

$$6a=-18$$

Divide both sides by 6.

$$a=\frac{-18}{6}$$

Divide -18 by 6 to get -3.

$$a=-3$$