Solve b^2+b(3-3)+3b=(b-3)^2

$$b ^ { 2 } + b ( 3 - 3 ) + 3 b = ( b - 3 ) ^ { 2 }$$

Solve for b

$$b=1$$

Steps for Solving Linear Equation

Subtract 3 from 3 to get 0.

$$b^{2}+b\times 0+3b=\left(b-3\right)^{2}$$

Anything times zero gives zero.

$$b^{2}+0+3b=\left(b-3\right)^{2}$$

Anything plus zero gives itself.

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

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

$$b^{2}+3b=b^{2}-6b+9$$

Subtract b^{2} from both sides.

$$b^{2}+3b-b^{2}=-6b+9$$

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

$$3b=-6b+9$$

Add 6b to both sides.

$$3b+6b=9$$

Combine 3b and 6b to get 9b.

$$9b=9$$

Divide both sides by 9.

$$b=\frac{9}{9}$$

Divide 9 by 9 to get 1.

$$b=1$$