Solve (x-4)^2+33=15b

$$( x - 4 ) ^ { 2 } + 33 = 15 b$$

Solve for b

$$b=\frac{\left(x-4\right)^{2}+33}{15}$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$x^{2}-8x+16+33=15b$$

Add 16 and 33 to get 49.

$$x^{2}-8x+49=15b$$

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

$$15b=x^{2}-8x+49$$

Divide both sides by 15.

$$\frac{15b}{15}=\frac{x^{2}-8x+49}{15}$$

Dividing by 15 undoes the multiplication by 15.

$$b=\frac{x^{2}-8x+49}{15}$$

Solve for x (complex solution)

$$x=-\sqrt{15b-33}+4$$

$$x=\sqrt{15b-33}+4$$

Solve for x

$$x=-\sqrt{15b-33}+4$$

$$x=\sqrt{15b-33}+4\text{, }b\geq \frac{11}{5}$$