Solve (2+sqrt{3})^2-(2sqrt{b}-x)^2=16-x^2

$$( 2 + \sqrt { 3 } ) ^ { 2 } - ( 2 \sqrt { b } - x ) ^ { 2 } = 16 - x ^ { 2 }$$

Solve for x (complex solution)

$$x=-\frac{b^{-\frac{1}{2}}\left(-4b+4\sqrt{3}-9\right)}{4}$$

$$b\neq 0$$

Steps for Solving Linear Equation

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

$$4+4\sqrt{3}+\left(\sqrt{3}\right)^{2}-\left(2\sqrt{b}-x\right)^{2}=16-x^{2}$$

The square of \sqrt{3} is 3.

$$4+4\sqrt{3}+3-\left(2\sqrt{b}-x\right)^{2}=16-x^{2}$$

Add 4 and 3 to get 7.

$$7+4\sqrt{3}-\left(2\sqrt{b}-x\right)^{2}=16-x^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(2\sqrt{b}-x\right)^{2}.

$$7+4\sqrt{3}-\left(4\left(\sqrt{b}\right)^{2}-4\sqrt{b}x+x^{2}\right)=16-x^{2}$$

Calculate \sqrt{b} to the power of 2 and get b.

$$7+4\sqrt{3}-\left(4b-4\sqrt{b}x+x^{2}\right)=16-x^{2}$$

To find the opposite of 4b-4\sqrt{b}x+x^{2}, find the opposite of each term.

$$7+4\sqrt{3}-4b+4\sqrt{b}x-x^{2}=16-x^{2}$$

Add x^{2} to both sides.

$$7+4\sqrt{3}-4b+4\sqrt{b}x-x^{2}+x^{2}=16$$

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

$$7+4\sqrt{3}-4b+4\sqrt{b}x=16$$

Subtract 7 from both sides.

$$4\sqrt{3}-4b+4\sqrt{b}x=16-7$$

Subtract 7 from 16 to get 9.

$$4\sqrt{3}-4b+4\sqrt{b}x=9$$

Subtract 4\sqrt{3} from both sides.

$$-4b+4\sqrt{b}x=9-4\sqrt{3}$$

Add 4b to both sides.

$$4\sqrt{b}x=9-4\sqrt{3}+4b$$

The equation is in standard form.

$$4\sqrt{b}x=4b+9-4\sqrt{3}$$

Divide both sides by 4\sqrt{b}.

$$\frac{4\sqrt{b}x}{4\sqrt{b}}=\frac{4b+9-4\sqrt{3}}{4\sqrt{b}}$$

Dividing by 4\sqrt{b} undoes the multiplication by 4\sqrt{b}.

$$x=\frac{4b+9-4\sqrt{3}}{4\sqrt{b}}$$

Divide 9-4\sqrt{3}+4b by 4\sqrt{b}.

$$x=\frac{b^{-\frac{1}{2}}\left(4b+9-4\sqrt{3}\right)}{4}$$

Solve for x

$$x=-\frac{-4b+4\sqrt{3}-9}{4\sqrt{b}}$$

$$b>0$$

Steps for Solving Linear Equation

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

$$4+4\sqrt{3}+\left(\sqrt{3}\right)^{2}-\left(2\sqrt{b}-x\right)^{2}=16-x^{2}$$

The square of \sqrt{3} is 3.

$$4+4\sqrt{3}+3-\left(2\sqrt{b}-x\right)^{2}=16-x^{2}$$

Add 4 and 3 to get 7.

$$7+4\sqrt{3}-\left(2\sqrt{b}-x\right)^{2}=16-x^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(2\sqrt{b}-x\right)^{2}.

$$7+4\sqrt{3}-\left(4\left(\sqrt{b}\right)^{2}-4\sqrt{b}x+x^{2}\right)=16-x^{2}$$

Calculate \sqrt{b} to the power of 2 and get b.

$$7+4\sqrt{3}-\left(4b-4\sqrt{b}x+x^{2}\right)=16-x^{2}$$

To find the opposite of 4b-4\sqrt{b}x+x^{2}, find the opposite of each term.

$$7+4\sqrt{3}-4b+4\sqrt{b}x-x^{2}=16-x^{2}$$

Add x^{2} to both sides.

$$7+4\sqrt{3}-4b+4\sqrt{b}x-x^{2}+x^{2}=16$$

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

$$7+4\sqrt{3}-4b+4\sqrt{b}x=16$$

Subtract 7 from both sides.

$$4\sqrt{3}-4b+4\sqrt{b}x=16-7$$

Subtract 7 from 16 to get 9.

$$4\sqrt{3}-4b+4\sqrt{b}x=9$$

Subtract 4\sqrt{3} from both sides.

$$-4b+4\sqrt{b}x=9-4\sqrt{3}$$

Add 4b to both sides.

$$4\sqrt{b}x=9-4\sqrt{3}+4b$$

The equation is in standard form.

$$4\sqrt{b}x=4b+9-4\sqrt{3}$$

Divide both sides by 4\sqrt{b}.

$$\frac{4\sqrt{b}x}{4\sqrt{b}}=\frac{4b+9-4\sqrt{3}}{4\sqrt{b}}$$

Dividing by 4\sqrt{b} undoes the multiplication by 4\sqrt{b}.

$$x=\frac{4b+9-4\sqrt{3}}{4\sqrt{b}}$$

Solve for b (complex solution)

$$\left\{\begin{matrix}b=\frac{\left(\sqrt{x^{2}+4\sqrt{3}-9}+x\right)^{2}}{4}\text{, }&arg(\frac{\sqrt{x^{2}+4\sqrt{3}-9}+x}{2})<\pi \\b=\frac{\left(-\sqrt{x^{2}+4\sqrt{3}-9}+x\right)^{2}}{4}\text{, }&arg(\frac{-\sqrt{x^{2}+4\sqrt{3}-9}+x}{2})<\pi \end{matrix}\right.$$

Solve for b

$$b=\frac{\left(\sqrt{x^{2}+4\sqrt{3}-9}+x\right)^{2}}{4}$$

$$b=\frac{\left(-\sqrt{x^{2}+4\sqrt{3}-9}+x\right)^{2}}{4}\text{, }x\geq \sqrt{9-4\sqrt{3}}$$