Solve (x+2)^2-(3-x)^2>5

$$( x + 2 ) ^ { 2 } - ( 3 - x ) ^ { 2 } > 5$$

Solve for x

$$x>1$$

Solution Steps

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

$$x^{2}+4x+4-\left(3-x\right)^{2}>5$$

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

$$x^{2}+4x+4-\left(9-6x+x^{2}\right)>5$$

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

$$x^{2}+4x+4-9+6x-x^{2}>5$$

Subtract 9 from 4 to get -5.

$$x^{2}+4x-5+6x-x^{2}>5$$

Combine 4x and 6x to get 10x.

$$x^{2}+10x-5-x^{2}>5$$

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

$$10x-5>5$$

Add 5 to both sides.

$$10x>5+5$$

Add 5 and 5 to get 10.

$$10x>10$$

Divide both sides by 10. Since 10 is positive, the inequality direction remains the same.

$$x>\frac{10}{10}$$

Divide 10 by 10 to get 1.

$$x>1$$