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

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

Solve for x

$$x=\frac{\sqrt{2}+\sqrt{6}}{4}\approx 0.965925826$$

Steps for Solving Linear Equation

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

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

Calculate 3 to the power of 2 and get 9.

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

The square of \sqrt{2} is 2.

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

Multiply 9 and 2 to get 18.

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

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

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

Calculate 2 to the power of 2 and get 4.

$$x=\frac{18+4\left(\sqrt{3}\right)^{2}-\left(3-\sqrt{3}\right)^{2}}{2\times 2\times 3\sqrt{2}\sqrt{3}}$$

The square of \sqrt{3} is 3.

$$x=\frac{18+4\times 3-\left(3-\sqrt{3}\right)^{2}}{2\times 2\times 3\sqrt{2}\sqrt{3}}$$

Multiply 4 and 3 to get 12.

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

Add 18 and 12 to get 30.

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

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

$$x=\frac{30-\left(9-6\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)}{2\times 2\times 3\sqrt{2}\sqrt{3}}$$

The square of \sqrt{3} is 3.

$$x=\frac{30-\left(9-6\sqrt{3}+3\right)}{2\times 2\times 3\sqrt{2}\sqrt{3}}$$

Add 9 and 3 to get 12.

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

To find the opposite of 12-6\sqrt{3}, find the opposite of each term.

$$x=\frac{30-12+6\sqrt{3}}{2\times 2\times 3\sqrt{2}\sqrt{3}}$$

Subtract 12 from 30 to get 18.

$$x=\frac{18+6\sqrt{3}}{2\times 2\times 3\sqrt{2}\sqrt{3}}$$

Multiply 2 and 2 to get 4.

$$x=\frac{18+6\sqrt{3}}{4\times 3\sqrt{2}\sqrt{3}}$$

Multiply 4 and 3 to get 12.

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

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

$$x=\frac{18+6\sqrt{3}}{12\sqrt{6}}$$

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

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

The square of \sqrt{6} is 6.

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

Multiply 12 and 6 to get 72.

$$x=\frac{\left(18+6\sqrt{3}\right)\sqrt{6}}{72}$$

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

$$x=\frac{18\sqrt{6}+6\sqrt{3}\sqrt{6}}{72}$$

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}.

$$x=\frac{18\sqrt{6}+6\sqrt{3}\sqrt{3}\sqrt{2}}{72}$$

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

$$x=\frac{18\sqrt{6}+6\times 3\sqrt{2}}{72}$$

Multiply 6 and 3 to get 18.

$$x=\frac{18\sqrt{6}+18\sqrt{2}}{72}$$

Divide each term of 18\sqrt{6}+18\sqrt{2} by 72 to get \frac{1}{4}\sqrt{6}+\frac{1}{4}\sqrt{2}.

$$x=\frac{1}{4}\sqrt{6}+\frac{1}{4}\sqrt{2}$$

Assign x

$$x≔\frac{\sqrt{2}+\sqrt{6}}{4}$$