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

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

Solve for x

$$x = \frac{9}{2} = 4\frac{1}{2} = 4.5$$

Steps for Solving Linear Equation

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

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

Calculate 3 to the power of 2 and get 9.

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

The square of \sqrt{3} is 3.

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

Multiply 9 and 3 to get 27.

$$27-x^{2}=3^{2}-\left(6-x\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$27-x^{2}=9-\left(6-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6-x\right)^{2}.

$$27-x^{2}=9-\left(36-12x+x^{2}\right)$$

To find the opposite of 36-12x+x^{2}, find the opposite of each term.

$$27-x^{2}=9-36+12x-x^{2}$$

Subtract 36 from 9 to get -27.

$$27-x^{2}=-27+12x-x^{2}$$

Subtract 12x from both sides.

$$27-x^{2}-12x=-27-x^{2}$$

Add x^{2} to both sides.

$$27-x^{2}-12x+x^{2}=-27$$

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

$$27-12x=-27$$

Subtract 27 from both sides.

$$-12x=-27-27$$

Subtract 27 from -27 to get -54.

$$-12x=-54$$

Divide both sides by -12.

$$x=\frac{-54}{-12}$$

Reduce the fraction \frac{-54}{-12} to lowest terms by extracting and canceling out -6.

$$x=\frac{9}{2}$$