Solve A=frac{2^2+(sqrt{3}-1)^2-(sqrt{2})^2}{2*2*(sqrt{3}-1)}

$$A = \frac { 2 ^ { 2 } + ( \sqrt { 3 } - 1 ) ^ { 2 } - ( \sqrt { 2 } ) ^ { 2 } } { 2 \cdot 2 \cdot ( \sqrt { 3 } - 1 ) }$$

Solve for A

$$A=\frac{\sqrt{3}}{2}\approx 0.866025404$$

Steps for Solving Linear Equation

Calculate 2 to the power of 2 and get 4.

$$A=\frac{4+\left(\sqrt{3}-1\right)^{2}-\left(\sqrt{2}\right)^{2}}{2\times 2\left(\sqrt{3}-1\right)}$$

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

$$A=\frac{4+\left(\sqrt{3}\right)^{2}-2\sqrt{3}+1-\left(\sqrt{2}\right)^{2}}{2\times 2\left(\sqrt{3}-1\right)}$$

The square of \sqrt{3} is 3.

$$A=\frac{4+3-2\sqrt{3}+1-\left(\sqrt{2}\right)^{2}}{2\times 2\left(\sqrt{3}-1\right)}$$

Add 3 and 1 to get 4.

$$A=\frac{4+4-2\sqrt{3}-\left(\sqrt{2}\right)^{2}}{2\times 2\left(\sqrt{3}-1\right)}$$

Add 4 and 4 to get 8.

$$A=\frac{8-2\sqrt{3}-\left(\sqrt{2}\right)^{2}}{2\times 2\left(\sqrt{3}-1\right)}$$

The square of \sqrt{2} is 2.

$$A=\frac{8-2\sqrt{3}-2}{2\times 2\left(\sqrt{3}-1\right)}$$

Subtract 2 from 8 to get 6.

$$A=\frac{6-2\sqrt{3}}{2\times 2\left(\sqrt{3}-1\right)}$$

Multiply 2 and 2 to get 4.

$$A=\frac{6-2\sqrt{3}}{4\left(\sqrt{3}-1\right)}$$

Use the distributive property to multiply 4 by \sqrt{3}-1.

$$A=\frac{6-2\sqrt{3}}{4\sqrt{3}-4}$$

Rationalize the denominator of \frac{6-2\sqrt{3}}{4\sqrt{3}-4} by multiplying numerator and denominator by 4\sqrt{3}+4.

$$A=\frac{\left(6-2\sqrt{3}\right)\left(4\sqrt{3}+4\right)}{\left(4\sqrt{3}-4\right)\left(4\sqrt{3}+4\right)}$$

Consider \left(4\sqrt{3}-4\right)\left(4\sqrt{3}+4\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$A=\frac{\left(6-2\sqrt{3}\right)\left(4\sqrt{3}+4\right)}{\left(4\sqrt{3}\right)^{2}-4^{2}}$$

Expand \left(4\sqrt{3}\right)^{2}.

$$A=\frac{\left(6-2\sqrt{3}\right)\left(4\sqrt{3}+4\right)}{4^{2}\left(\sqrt{3}\right)^{2}-4^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$A=\frac{\left(6-2\sqrt{3}\right)\left(4\sqrt{3}+4\right)}{16\left(\sqrt{3}\right)^{2}-4^{2}}$$

The square of \sqrt{3} is 3.

$$A=\frac{\left(6-2\sqrt{3}\right)\left(4\sqrt{3}+4\right)}{16\times 3-4^{2}}$$

Multiply 16 and 3 to get 48.

$$A=\frac{\left(6-2\sqrt{3}\right)\left(4\sqrt{3}+4\right)}{48-4^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$A=\frac{\left(6-2\sqrt{3}\right)\left(4\sqrt{3}+4\right)}{48-16}$$

Subtract 16 from 48 to get 32.

$$A=\frac{\left(6-2\sqrt{3}\right)\left(4\sqrt{3}+4\right)}{32}$$

Use the distributive property to multiply 6-2\sqrt{3} by 4\sqrt{3}+4 and combine like terms.

$$A=\frac{16\sqrt{3}+24-8\left(\sqrt{3}\right)^{2}}{32}$$

The square of \sqrt{3} is 3.

$$A=\frac{16\sqrt{3}+24-8\times 3}{32}$$

Multiply -8 and 3 to get -24.

$$A=\frac{16\sqrt{3}+24-24}{32}$$

Subtract 24 from 24 to get 0.

$$A=\frac{16\sqrt{3}}{32}$$

Divide 16\sqrt{3} by 32 to get \frac{1}{2}\sqrt{3}.

$$A=\frac{1}{2}\sqrt{3}$$

Assign A

$$A≔\frac{\sqrt{3}}{2}$$