Solve 12=(1-3x)(1-3x)d+(1+3x)(1+3x)

$$12 = ( 1 - 3 x ) ( 1 - 3 x ) d + ( 1 + 3 x ) ( 1 + 3 x )$$

Solve for d

$$d=-\frac{9x^{2}+6x-11}{\left(1-3x\right)^{2}}$$

$$x\neq \frac{1}{3}$$

Steps for Solving Linear Equation

Multiply 1-3x and 1-3x to get \left(1-3x\right)^{2}.

$$12=\left(1-3x\right)^{2}d+\left(1+3x\right)\left(1+3x\right)$$

Multiply 1+3x and 1+3x to get \left(1+3x\right)^{2}.

$$12=\left(1-3x\right)^{2}d+\left(1+3x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-3x\right)^{2}.

$$12=\left(1-6x+9x^{2}\right)d+\left(1+3x\right)^{2}$$

Use the distributive property to multiply 1-6x+9x^{2} by d.

$$12=d-6xd+9x^{2}d+\left(1+3x\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+3x\right)^{2}.

$$12=d-6xd+9x^{2}d+1+6x+9x^{2}$$

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

$$d-6xd+9x^{2}d+1+6x+9x^{2}=12$$

Subtract 1 from both sides.

$$d-6xd+9x^{2}d+6x+9x^{2}=12-1$$

Subtract 1 from 12 to get 11.

$$d-6xd+9x^{2}d+6x+9x^{2}=11$$

Subtract 6x from both sides.

$$d-6xd+9x^{2}d+9x^{2}=11-6x$$

Subtract 9x^{2} from both sides.

$$d-6xd+9x^{2}d=11-6x-9x^{2}$$

Combine all terms containing d.

$$\left(1-6x+9x^{2}\right)d=11-6x-9x^{2}$$

The equation is in standard form.

$$\left(9x^{2}-6x+1\right)d=11-6x-9x^{2}$$

Divide both sides by 1-6x+9x^{2}.

$$\frac{\left(9x^{2}-6x+1\right)d}{9x^{2}-6x+1}=\frac{11-6x-9x^{2}}{9x^{2}-6x+1}$$

Dividing by 1-6x+9x^{2} undoes the multiplication by 1-6x+9x^{2}.

$$d=\frac{11-6x-9x^{2}}{9x^{2}-6x+1}$$

Divide 11-6x-9x^{2} by 1-6x+9x^{2}.

$$d=\frac{11-6x-9x^{2}}{\left(3x-1\right)^{2}}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=-\frac{-d+2\sqrt{2d+3}+1}{3\left(d+1\right)}\text{; }x=-\frac{-d-2\sqrt{2d+3}+1}{3\left(d+1\right)}\text{, }&d\neq -1\\x=1\text{, }&d=-1\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=-\frac{-d+2\sqrt{2d+3}+1}{3\left(d+1\right)}\text{; }x=-\frac{-d-2\sqrt{2d+3}+1}{3\left(d+1\right)}\text{, }&d\neq -1\text{ and }d\geq -\frac{3}{2}\\x=1\text{, }&d=-1\end{matrix}\right.$$