Solve 9x-7(x+3)+(x+3)^2+x+3quad1=x^2

$$9 x - 7 ( x + 3 ) + ( x + 3 ) ^ { 2 } + x + 3 \quad 1 = x ^ { 2 }$$

Solve for x

$$x=1$$

Steps for Solving Linear Equation

Use the distributive property to multiply -7 by x+3.

$$9x-7x-21+\left(x+3\right)^{2}+x+3\times 1=x^{2}$$

Combine 9x and -7x to get 2x.

$$2x-21+\left(x+3\right)^{2}+x+3\times 1=x^{2}$$

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

$$2x-21+x^{2}+6x+9+x+3\times 1=x^{2}$$

Combine 2x and 6x to get 8x.

$$8x-21+x^{2}+9+x+3\times 1=x^{2}$$

Add -21 and 9 to get -12.

$$8x-12+x^{2}+x+3\times 1=x^{2}$$

Combine 8x and x to get 9x.

$$9x-12+x^{2}+3\times 1=x^{2}$$

Multiply 3 and 1 to get 3.

$$9x-12+x^{2}+3=x^{2}$$

Add -12 and 3 to get -9.

$$9x-9+x^{2}=x^{2}$$

Subtract x^{2} from both sides.

$$9x-9+x^{2}-x^{2}=0$$

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

$$9x-9=0$$

Add 9 to both sides. Anything plus zero gives itself.

$$9x=9$$

Divide both sides by 9.

$$x=\frac{9}{9}$$

Divide 9 by 9 to get 1.

$$x=1$$