Solve (3-x)(3+x)=3-(x-3)^2

$$( 3 - x ) ( 3 + x ) = 3 - ( x - 3 ) ^ { 2 }$$

Solve for x

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

Steps for Solving Linear Equation

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

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

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

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

To find the opposite of x^{2}-6x+9, find the opposite of each term.

$$9-x^{2}=3-x^{2}+6x-9$$

Subtract 9 from 3 to get -6.

$$9-x^{2}=-6-x^{2}+6x$$

Add x^{2} to both sides.

$$9-x^{2}+x^{2}=-6+6x$$

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

$$9=-6+6x$$

Swap sides so that all variable terms are on the left hand side.

$$-6+6x=9$$

Add 6 to both sides.

$$6x=9+6$$

Add 9 and 6 to get 15.

$$6x=15$$

Divide both sides by 6.

$$x=\frac{15}{6}$$

Reduce the fraction \frac{15}{6} to lowest terms by extracting and canceling out 3.

$$x=\frac{5}{2}$$