Solve left(1-xright)^2+1/4=x^2

$${ \left(1-x \right) }^{ 2 } + \frac{ 1 }{ 4 } = { x }^{ 2 }$$

Solve for x

$$x=\frac{5}{8}=0.625$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-x\right)^{2}.

$$1-2x+x^{2}+\frac{1}{4}=x^{2}$$

Add 1 and \frac{1}{4} to get \frac{5}{4}.

$$\frac{5}{4}-2x+x^{2}=x^{2}$$

Subtract x^{2} from both sides.

$$\frac{5}{4}-2x+x^{2}-x^{2}=0$$

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

$$\frac{5}{4}-2x=0$$

Subtract \frac{5}{4} from both sides. Anything subtracted from zero gives its negation.

$$-2x=-\frac{5}{4}$$

Divide both sides by -2.

$$x=\frac{-\frac{5}{4}}{-2}$$

Express \frac{-\frac{5}{4}}{-2} as a single fraction.

$$x=\frac{-5}{4\left(-2\right)}$$

Multiply 4 and -2 to get -8.

$$x=\frac{-5}{-8}$$

Fraction \frac{-5}{-8} can be simplified to \frac{5}{8} by removing the negative sign from both the numerator and the denominator.

$$x=\frac{5}{8}$$