Solve x=frac{{left(3sqrt{2}right)}^2+{left(3-sqrt{3}right)}^2-{left(2sqrt{3}right)}^2}{2*3sqrt{2}left(3-sqrt{3}right)}

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

Solve for x

$$x=\frac{\sqrt{2}}{2}\approx 0.707106781$$

Steps for Solving Linear Equation

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

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

Calculate 3 to the power of 2 and get 9.

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

The square of \sqrt{2} is 2.

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

Multiply 9 and 2 to get 18.

$$x=\frac{18+\left(3-\sqrt{3}\right)^{2}-\left(2\sqrt{3}\right)^{2}}{2\times 3\sqrt{2}\left(3-\sqrt{3}\right)}$$

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

$$x=\frac{18+9-6\sqrt{3}+\left(\sqrt{3}\right)^{2}-\left(2\sqrt{3}\right)^{2}}{2\times 3\sqrt{2}\left(3-\sqrt{3}\right)}$$

The square of \sqrt{3} is 3.

$$x=\frac{18+9-6\sqrt{3}+3-\left(2\sqrt{3}\right)^{2}}{2\times 3\sqrt{2}\left(3-\sqrt{3}\right)}$$

Add 9 and 3 to get 12.

$$x=\frac{18+12-6\sqrt{3}-\left(2\sqrt{3}\right)^{2}}{2\times 3\sqrt{2}\left(3-\sqrt{3}\right)}$$

Add 18 and 12 to get 30.

$$x=\frac{30-6\sqrt{3}-\left(2\sqrt{3}\right)^{2}}{2\times 3\sqrt{2}\left(3-\sqrt{3}\right)}$$

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

$$x=\frac{30-6\sqrt{3}-2^{2}\left(\sqrt{3}\right)^{2}}{2\times 3\sqrt{2}\left(3-\sqrt{3}\right)}$$

Calculate 2 to the power of 2 and get 4.

$$x=\frac{30-6\sqrt{3}-4\left(\sqrt{3}\right)^{2}}{2\times 3\sqrt{2}\left(3-\sqrt{3}\right)}$$

The square of \sqrt{3} is 3.

$$x=\frac{30-6\sqrt{3}-4\times 3}{2\times 3\sqrt{2}\left(3-\sqrt{3}\right)}$$

Multiply 4 and 3 to get 12.

$$x=\frac{30-6\sqrt{3}-12}{2\times 3\sqrt{2}\left(3-\sqrt{3}\right)}$$

Subtract 12 from 30 to get 18.

$$x=\frac{18-6\sqrt{3}}{2\times 3\sqrt{2}\left(3-\sqrt{3}\right)}$$

Multiply 2 and 3 to get 6.

$$x=\frac{18-6\sqrt{3}}{6\sqrt{2}\left(3-\sqrt{3}\right)}$$

Rationalize the denominator of \frac{18-6\sqrt{3}}{6\sqrt{2}\left(3-\sqrt{3}\right)} by multiplying numerator and denominator by \sqrt{2}.

$$x=\frac{\left(18-6\sqrt{3}\right)\sqrt{2}}{6\left(\sqrt{2}\right)^{2}\left(3-\sqrt{3}\right)}$$

The square of \sqrt{2} is 2.

$$x=\frac{\left(18-6\sqrt{3}\right)\sqrt{2}}{6\times 2\left(3-\sqrt{3}\right)}$$

Multiply 6 and 2 to get 12.

$$x=\frac{\left(18-6\sqrt{3}\right)\sqrt{2}}{12\left(3-\sqrt{3}\right)}$$

Use the distributive property to multiply 18-6\sqrt{3} by \sqrt{2}.

$$x=\frac{18\sqrt{2}-6\sqrt{3}\sqrt{2}}{12\left(3-\sqrt{3}\right)}$$

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

$$x=\frac{18\sqrt{2}-6\sqrt{6}}{12\left(3-\sqrt{3}\right)}$$

Use the distributive property to multiply 12 by 3-\sqrt{3}.

$$x=\frac{18\sqrt{2}-6\sqrt{6}}{36-12\sqrt{3}}$$

Rationalize the denominator of \frac{18\sqrt{2}-6\sqrt{6}}{36-12\sqrt{3}} by multiplying numerator and denominator by 36+12\sqrt{3}.

$$x=\frac{\left(18\sqrt{2}-6\sqrt{6}\right)\left(36+12\sqrt{3}\right)}{\left(36-12\sqrt{3}\right)\left(36+12\sqrt{3}\right)}$$

Consider \left(36-12\sqrt{3}\right)\left(36+12\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$x=\frac{\left(18\sqrt{2}-6\sqrt{6}\right)\left(36+12\sqrt{3}\right)}{36^{2}-\left(-12\sqrt{3}\right)^{2}}$$

Calculate 36 to the power of 2 and get 1296.

$$x=\frac{\left(18\sqrt{2}-6\sqrt{6}\right)\left(36+12\sqrt{3}\right)}{1296-\left(-12\sqrt{3}\right)^{2}}$$

Expand \left(-12\sqrt{3}\right)^{2}.

$$x=\frac{\left(18\sqrt{2}-6\sqrt{6}\right)\left(36+12\sqrt{3}\right)}{1296-\left(-12\right)^{2}\left(\sqrt{3}\right)^{2}}$$

Calculate -12 to the power of 2 and get 144.

$$x=\frac{\left(18\sqrt{2}-6\sqrt{6}\right)\left(36+12\sqrt{3}\right)}{1296-144\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$x=\frac{\left(18\sqrt{2}-6\sqrt{6}\right)\left(36+12\sqrt{3}\right)}{1296-144\times 3}$$

Multiply 144 and 3 to get 432.

$$x=\frac{\left(18\sqrt{2}-6\sqrt{6}\right)\left(36+12\sqrt{3}\right)}{1296-432}$$

Subtract 432 from 1296 to get 864.

$$x=\frac{\left(18\sqrt{2}-6\sqrt{6}\right)\left(36+12\sqrt{3}\right)}{864}$$

Use the distributive property to multiply 18\sqrt{2}-6\sqrt{6} by 36+12\sqrt{3}.

$$x=\frac{648\sqrt{2}+216\sqrt{3}\sqrt{2}-216\sqrt{6}-72\sqrt{3}\sqrt{6}}{864}$$

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

$$x=\frac{648\sqrt{2}+216\sqrt{6}-216\sqrt{6}-72\sqrt{3}\sqrt{6}}{864}$$

Combine 216\sqrt{6} and -216\sqrt{6} to get 0.

$$x=\frac{648\sqrt{2}-72\sqrt{3}\sqrt{6}}{864}$$

Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.

$$x=\frac{648\sqrt{2}-72\sqrt{3}\sqrt{3}\sqrt{2}}{864}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$x=\frac{648\sqrt{2}-72\times 3\sqrt{2}}{864}$$

Multiply -72 and 3 to get -216.

$$x=\frac{648\sqrt{2}-216\sqrt{2}}{864}$$

Combine 648\sqrt{2} and -216\sqrt{2} to get 432\sqrt{2}.

$$x=\frac{432\sqrt{2}}{864}$$

Divide 432\sqrt{2} by 864 to get \frac{1}{2}\sqrt{2}.

$$x=\frac{1}{2}\sqrt{2}$$

Assign x

$$x≔\frac{\sqrt{2}}{2}$$