Solve C=(sqrt{3}-sqrt{2})^2+3sqrt{6}

$$C = ( \sqrt { 3 } - \sqrt { 2 } ) ^ { 2 } + 3 \sqrt { 6 }$$

Solve for C

$$C=\sqrt{6}+5\approx 7.449489743$$

Steps for Solving Linear Equation

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

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

The square of \sqrt{3} is 3.

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

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

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

The square of \sqrt{2} is 2.

$$C=3-2\sqrt{6}+2+3\sqrt{6}$$

Add 3 and 2 to get 5.

$$C=5-2\sqrt{6}+3\sqrt{6}$$

Combine -2\sqrt{6} and 3\sqrt{6} to get \sqrt{6}.

$$C=5+\sqrt{6}$$

Assign C

$$C≔\sqrt{6}+5$$