Solve 3x^2+2-{left(sqrt{3}x-sqrt{2}right)}^2=0

$$3 { x }^{ 2 } +2- { \left( \sqrt{ 3 } x- \sqrt{ 2 } \right) }^{ 2 } = 0$$

Solve for x

$$x=0$$

Steps for Solving Linear Equation

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

$$3x^{2}+2-\left(\left(\sqrt{3}\right)^{2}x^{2}-2\sqrt{3}x\sqrt{2}+\left(\sqrt{2}\right)^{2}\right)=0$$

The square of \sqrt{3} is 3.

$$3x^{2}+2-\left(3x^{2}-2\sqrt{3}x\sqrt{2}+\left(\sqrt{2}\right)^{2}\right)=0$$

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

$$3x^{2}+2-\left(3x^{2}-2\sqrt{6}x+\left(\sqrt{2}\right)^{2}\right)=0$$

The square of \sqrt{2} is 2.

$$3x^{2}+2-\left(3x^{2}-2\sqrt{6}x+2\right)=0$$

To find the opposite of 3x^{2}-2\sqrt{6}x+2, find the opposite of each term.

$$3x^{2}+2-3x^{2}+2\sqrt{6}x-2=0$$

Combine 3x^{2} and -3x^{2} to get 0.

$$2+2\sqrt{6}x-2=0$$

Subtract 2 from 2 to get 0.

$$2\sqrt{6}x=0$$

Divide 0 by 2\sqrt{6}.

$$x=0$$