Solve d^2+(-d+6)^2=b(-d+6-1/3d^2)

$$d ^ { 2 } + ( - d + 6 ) ^ { 2 } = b ( - d + 6 - \frac { 1 } { 3 } d ^ { 2 } )$$

Solve for b

$$b=-\frac{6\left(d^{2}-6d+18\right)}{\left(d-3\right)\left(d+6\right)}$$

$$d\neq -6\text{ and }d\neq 3$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-d+6\right)^{2}.

$$d^{2}+\left(-d\right)^{2}+12\left(-d\right)+36=b\left(-d+6-\frac{1}{3}d^{2}\right)$$

Calculate -d to the power of 2 and get d^{2}.

$$d^{2}+d^{2}+12\left(-d\right)+36=b\left(-d+6-\frac{1}{3}d^{2}\right)$$

Combine d^{2} and d^{2} to get 2d^{2}.

$$2d^{2}+12\left(-d\right)+36=b\left(-d+6-\frac{1}{3}d^{2}\right)$$

Use the distributive property to multiply b by -d+6-\frac{1}{3}d^{2}.

$$2d^{2}+12\left(-d\right)+36=b\left(-d\right)+6b-\frac{1}{3}bd^{2}$$

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

$$b\left(-d\right)+6b-\frac{1}{3}bd^{2}=2d^{2}+12\left(-d\right)+36$$

Multiply 12 and -1 to get -12.

$$b\left(-1\right)d+6b-\frac{1}{3}bd^{2}=2d^{2}-12d+36$$

Combine all terms containing b.

$$\left(-d+6-\frac{1}{3}d^{2}\right)b=2d^{2}-12d+36$$

The equation is in standard form.

$$\left(-\frac{d^{2}}{3}-d+6\right)b=2d^{2}-12d+36$$

Divide both sides by -d+6-\frac{1}{3}d^{2}.

$$\frac{\left(-\frac{d^{2}}{3}-d+6\right)b}{-\frac{d^{2}}{3}-d+6}=\frac{2d^{2}-12d+36}{-\frac{d^{2}}{3}-d+6}$$

Dividing by -d+6-\frac{1}{3}d^{2} undoes the multiplication by -d+6-\frac{1}{3}d^{2}.

$$b=\frac{2d^{2}-12d+36}{-\frac{d^{2}}{3}-d+6}$$

Divide 2d^{2}+36-12d by -d+6-\frac{1}{3}d^{2}.

$$b=\frac{6\left(d^{2}-6d+18\right)}{\left(3-d\right)\left(d+6\right)}$$

Solve for d (complex solution)

$$\left\{\begin{matrix}d=\frac{3\left(\sqrt{3\left(3b^{2}-8b-48\right)}-b+12\right)}{2\left(b+6\right)}\text{; }d=\frac{3\left(-\sqrt{3\left(3b^{2}-8b-48\right)}-b+12\right)}{2\left(b+6\right)}\text{, }&b\neq -6\\d=4\text{, }&b=-6\end{matrix}\right.$$

Solve for d

$$\left\{\begin{matrix}d=\frac{3\left(\sqrt{3\left(3b^{2}-8b-48\right)}-b+12\right)}{2\left(b+6\right)}\text{; }d=\frac{3\left(-\sqrt{3\left(3b^{2}-8b-48\right)}-b+12\right)}{2\left(b+6\right)}\text{, }&\left(b\neq -6\text{ and }b\leq \frac{4-4\sqrt{10}}{3}\right)\text{ or }b\geq \frac{4\sqrt{10}+4}{3}\\d=4\text{, }&b=-6\end{matrix}\right.$$