Solve 16+left(4+xright)^2=left(9-xright)^2

$$16+ { \left(4+x \right) }^{ 2 } = { \left(9-x \right) }^{ 2 }$$

Solve for x

$$x = \frac{49}{26} = 1\frac{23}{26} \approx 1.884615385$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4+x\right)^{2}.

$$16+16+8x+x^{2}=\left(9-x\right)^{2}$$

Add 16 and 16 to get 32.

$$32+8x+x^{2}=\left(9-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(9-x\right)^{2}.

$$32+8x+x^{2}=81-18x+x^{2}$$

Add 18x to both sides.

$$32+8x+x^{2}+18x=81+x^{2}$$

Combine 8x and 18x to get 26x.

$$32+26x+x^{2}=81+x^{2}$$

Subtract x^{2} from both sides.

$$32+26x+x^{2}-x^{2}=81$$

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

$$32+26x=81$$

Subtract 32 from both sides.

$$26x=81-32$$

Subtract 32 from 81 to get 49.

$$26x=49$$

Divide both sides by 26.

$$x=\frac{49}{26}$$