Solve N(17x)^2-(5x)^2=24

$$N ( 17 x ) ^ { 2 } - ( 5 x ) ^ { 2 } = 24$$

Solve for N

$$N=\frac{25}{289}+\frac{24}{289x^{2}}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Expand \left(17x\right)^{2}.

$$N\times 17^{2}x^{2}-\left(5x\right)^{2}=24$$

Calculate 17 to the power of 2 and get 289.

$$N\times 289x^{2}-\left(5x\right)^{2}=24$$

Expand \left(5x\right)^{2}.

$$N\times 289x^{2}-5^{2}x^{2}=24$$

Calculate 5 to the power of 2 and get 25.

$$N\times 289x^{2}-25x^{2}=24$$

Add 25x^{2} to both sides.

$$N\times 289x^{2}=24+25x^{2}$$

The equation is in standard form.

$$289x^{2}N=25x^{2}+24$$

Divide both sides by 289x^{2}.

$$\frac{289x^{2}N}{289x^{2}}=\frac{25x^{2}+24}{289x^{2}}$$

Dividing by 289x^{2} undoes the multiplication by 289x^{2}.

$$N=\frac{25x^{2}+24}{289x^{2}}$$

Divide 24+25x^{2} by 289x^{2}.

$$N=\frac{25}{289}+\frac{24}{289x^{2}}$$

Solve for x (complex solution)

$$x=-2\sqrt{6}\left(289N-25\right)^{-\frac{1}{2}}$$

$$x=2\sqrt{6}\left(289N-25\right)^{-\frac{1}{2}}\text{, }N\neq \frac{25}{289}$$

Solve for x

$$x=2\sqrt{\frac{6}{289N-25}}$$

$$x=-2\sqrt{\frac{6}{289N-25}}\text{, }N>\frac{25}{289}$$