Solve (x-12/5)^2+(18/5)^2=x^2

$$( x - \frac { 12 } { 5 } ) ^ { 2 } + ( \frac { 18 } { 5 } ) ^ { 2 } = x ^ { 2 }$$

Solve for x

$$x = \frac{39}{10} = 3\frac{9}{10} = 3.9$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-\frac{12}{5}\right)^{2}.

$$x^{2}-\frac{24}{5}x+\frac{144}{25}+\left(\frac{18}{5}\right)^{2}=x^{2}$$

Calculate \frac{18}{5} to the power of 2 and get \frac{324}{25}.

$$x^{2}-\frac{24}{5}x+\frac{144}{25}+\frac{324}{25}=x^{2}$$

Add \frac{144}{25} and \frac{324}{25} to get \frac{468}{25}.

$$x^{2}-\frac{24}{5}x+\frac{468}{25}=x^{2}$$

Subtract x^{2} from both sides.

$$x^{2}-\frac{24}{5}x+\frac{468}{25}-x^{2}=0$$

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

$$-\frac{24}{5}x+\frac{468}{25}=0$$

Subtract \frac{468}{25} from both sides. Anything subtracted from zero gives its negation.

$$-\frac{24}{5}x=-\frac{468}{25}$$

Multiply both sides by -\frac{5}{24}, the reciprocal of -\frac{24}{5}.

$$x=-\frac{468}{25}\left(-\frac{5}{24}\right)$$

Multiply -\frac{468}{25} and -\frac{5}{24} to get \frac{39}{10}.

$$x=\frac{39}{10}$$