Solve (1/2-x)*x=[2/7(1-1/5)]:[(1-3/5):(1+frac{2}{5})]

$$( \frac { 1 } { 2 } - x ) \cdot x = [ \frac { 2 } { 7 } ( 1 - \frac { 1 } { 5 } ) ] : [ ( 1 - \frac { 3 } { 5 } ) : ( 1 + \frac { 2 } { 5 } ) ]$$

Solve for x (complex solution)

$$x=\frac{\sqrt{295}i}{20}+\frac{1}{4}\approx 0.25+0.858778202i$$

$$x=-\frac{\sqrt{295}i}{20}+\frac{1}{4}\approx 0.25-0.858778202i$$

Steps Using the Quadratic Formula

Use the distributive property to multiply \frac{1}{2}-x by x.

$$\frac{1}{2}x-x^{2}=\frac{\frac{2}{7}\left(1-\frac{1}{5}\right)}{\frac{1-\frac{3}{5}}{1+\frac{2}{5}}}$$

Convert 1 to fraction \frac{5}{5}.

$$\frac{1}{2}x-x^{2}=\frac{\frac{2}{7}\left(\frac{5}{5}-\frac{1}{5}\right)}{\frac{1-\frac{3}{5}}{1+\frac{2}{5}}}$$

Since \frac{5}{5} and \frac{1}{5} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1}{2}x-x^{2}=\frac{\frac{2}{7}\times \frac{5-1}{5}}{\frac{1-\frac{3}{5}}{1+\frac{2}{5}}}$$

Subtract 1 from 5 to get 4.

$$\frac{1}{2}x-x^{2}=\frac{\frac{2}{7}\times \frac{4}{5}}{\frac{1-\frac{3}{5}}{1+\frac{2}{5}}}$$

Multiply \frac{2}{7} times \frac{4}{5} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1}{2}x-x^{2}=\frac{\frac{2\times 4}{7\times 5}}{\frac{1-\frac{3}{5}}{1+\frac{2}{5}}}$$

Do the multiplications in the fraction \frac{2\times 4}{7\times 5}.

$$\frac{1}{2}x-x^{2}=\frac{\frac{8}{35}}{\frac{1-\frac{3}{5}}{1+\frac{2}{5}}}$$

Convert 1 to fraction \frac{5}{5}.

$$\frac{1}{2}x-x^{2}=\frac{\frac{8}{35}}{\frac{\frac{5}{5}-\frac{3}{5}}{1+\frac{2}{5}}}$$

Since \frac{5}{5} and \frac{3}{5} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1}{2}x-x^{2}=\frac{\frac{8}{35}}{\frac{\frac{5-3}{5}}{1+\frac{2}{5}}}$$

Subtract 3 from 5 to get 2.

$$\frac{1}{2}x-x^{2}=\frac{\frac{8}{35}}{\frac{\frac{2}{5}}{1+\frac{2}{5}}}$$

Convert 1 to fraction \frac{5}{5}.

$$\frac{1}{2}x-x^{2}=\frac{\frac{8}{35}}{\frac{\frac{2}{5}}{\frac{5}{5}+\frac{2}{5}}}$$

Since \frac{5}{5} and \frac{2}{5} have the same denominator, add them by adding their numerators.

$$\frac{1}{2}x-x^{2}=\frac{\frac{8}{35}}{\frac{\frac{2}{5}}{\frac{5+2}{5}}}$$

Add 5 and 2 to get 7.

$$\frac{1}{2}x-x^{2}=\frac{\frac{8}{35}}{\frac{\frac{2}{5}}{\frac{7}{5}}}$$

Divide \frac{2}{5} by \frac{7}{5} by multiplying \frac{2}{5} by the reciprocal of \frac{7}{5}.

$$\frac{1}{2}x-x^{2}=\frac{\frac{8}{35}}{\frac{2}{5}\times \frac{5}{7}}$$

Multiply \frac{2}{5} times \frac{5}{7} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1}{2}x-x^{2}=\frac{\frac{8}{35}}{\frac{2\times 5}{5\times 7}}$$

Cancel out 5 in both numerator and denominator.

$$\frac{1}{2}x-x^{2}=\frac{\frac{8}{35}}{\frac{2}{7}}$$

Divide \frac{8}{35} by \frac{2}{7} by multiplying \frac{8}{35} by the reciprocal of \frac{2}{7}.

$$\frac{1}{2}x-x^{2}=\frac{8}{35}\times \frac{7}{2}$$

Multiply \frac{8}{35} times \frac{7}{2} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1}{2}x-x^{2}=\frac{8\times 7}{35\times 2}$$

Do the multiplications in the fraction \frac{8\times 7}{35\times 2}.

$$\frac{1}{2}x-x^{2}=\frac{56}{70}$$

Reduce the fraction \frac{56}{70} to lowest terms by extracting and canceling out 14.

$$\frac{1}{2}x-x^{2}=\frac{4}{5}$$

Subtract \frac{4}{5} from both sides.

$$\frac{1}{2}x-x^{2}-\frac{4}{5}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-x^{2}+\frac{1}{2}x-\frac{4}{5}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, \frac{1}{2} for b, and -\frac{4}{5} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{1}{2}±\sqrt{\left(\frac{1}{2}\right)^{2}-4\left(-1\right)\left(-\frac{4}{5}\right)}}{2\left(-1\right)}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{1}{2}±\sqrt{\frac{1}{4}-4\left(-1\right)\left(-\frac{4}{5}\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\frac{1}{2}±\sqrt{\frac{1}{4}+4\left(-\frac{4}{5}\right)}}{2\left(-1\right)}$$

Multiply 4 times -\frac{4}{5}.

$$x=\frac{-\frac{1}{2}±\sqrt{\frac{1}{4}-\frac{16}{5}}}{2\left(-1\right)}$$

Add \frac{1}{4} to -\frac{16}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{1}{2}±\sqrt{-\frac{59}{20}}}{2\left(-1\right)}$$

Take the square root of -\frac{59}{20}.

$$x=\frac{-\frac{1}{2}±\frac{\sqrt{295}i}{10}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-\frac{1}{2}±\frac{\sqrt{295}i}{10}}{-2}$$

Now solve the equation x=\frac{-\frac{1}{2}±\frac{\sqrt{295}i}{10}}{-2} when ± is plus. Add -\frac{1}{2} to \frac{i\sqrt{295}}{10}.

$$x=\frac{\frac{\sqrt{295}i}{10}-\frac{1}{2}}{-2}$$

Divide -\frac{1}{2}+\frac{i\sqrt{295}}{10} by -2.

$$x=-\frac{\sqrt{295}i}{20}+\frac{1}{4}$$

Now solve the equation x=\frac{-\frac{1}{2}±\frac{\sqrt{295}i}{10}}{-2} when ± is minus. Subtract \frac{i\sqrt{295}}{10} from -\frac{1}{2}.

$$x=\frac{-\frac{\sqrt{295}i}{10}-\frac{1}{2}}{-2}$$

Divide -\frac{1}{2}-\frac{i\sqrt{295}}{10} by -2.

$$x=\frac{\sqrt{295}i}{20}+\frac{1}{4}$$

The equation is now solved.

$$x=-\frac{\sqrt{295}i}{20}+\frac{1}{4} x=\frac{\sqrt{295}i}{20}+\frac{1}{4}$$

Steps for Completing the Square

Use the distributive property to multiply \frac{1}{2}-x by x.

$$\frac{1}{2}x-x^{2}=\frac{\frac{2}{7}\left(1-\frac{1}{5}\right)}{\frac{1-\frac{3}{5}}{1+\frac{2}{5}}}$$

Convert 1 to fraction \frac{5}{5}.

$$\frac{1}{2}x-x^{2}=\frac{\frac{2}{7}\left(\frac{5}{5}-\frac{1}{5}\right)}{\frac{1-\frac{3}{5}}{1+\frac{2}{5}}}$$

Since \frac{5}{5} and \frac{1}{5} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1}{2}x-x^{2}=\frac{\frac{2}{7}\times \frac{5-1}{5}}{\frac{1-\frac{3}{5}}{1+\frac{2}{5}}}$$

Subtract 1 from 5 to get 4.

$$\frac{1}{2}x-x^{2}=\frac{\frac{2}{7}\times \frac{4}{5}}{\frac{1-\frac{3}{5}}{1+\frac{2}{5}}}$$

Multiply \frac{2}{7} times \frac{4}{5} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1}{2}x-x^{2}=\frac{\frac{2\times 4}{7\times 5}}{\frac{1-\frac{3}{5}}{1+\frac{2}{5}}}$$

Do the multiplications in the fraction \frac{2\times 4}{7\times 5}.

$$\frac{1}{2}x-x^{2}=\frac{\frac{8}{35}}{\frac{1-\frac{3}{5}}{1+\frac{2}{5}}}$$

Convert 1 to fraction \frac{5}{5}.

$$\frac{1}{2}x-x^{2}=\frac{\frac{8}{35}}{\frac{\frac{5}{5}-\frac{3}{5}}{1+\frac{2}{5}}}$$

Since \frac{5}{5} and \frac{3}{5} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1}{2}x-x^{2}=\frac{\frac{8}{35}}{\frac{\frac{5-3}{5}}{1+\frac{2}{5}}}$$

Subtract 3 from 5 to get 2.

$$\frac{1}{2}x-x^{2}=\frac{\frac{8}{35}}{\frac{\frac{2}{5}}{1+\frac{2}{5}}}$$

Convert 1 to fraction \frac{5}{5}.

$$\frac{1}{2}x-x^{2}=\frac{\frac{8}{35}}{\frac{\frac{2}{5}}{\frac{5}{5}+\frac{2}{5}}}$$

Since \frac{5}{5} and \frac{2}{5} have the same denominator, add them by adding their numerators.

$$\frac{1}{2}x-x^{2}=\frac{\frac{8}{35}}{\frac{\frac{2}{5}}{\frac{5+2}{5}}}$$

Add 5 and 2 to get 7.

$$\frac{1}{2}x-x^{2}=\frac{\frac{8}{35}}{\frac{\frac{2}{5}}{\frac{7}{5}}}$$

Divide \frac{2}{5} by \frac{7}{5} by multiplying \frac{2}{5} by the reciprocal of \frac{7}{5}.

$$\frac{1}{2}x-x^{2}=\frac{\frac{8}{35}}{\frac{2}{5}\times \frac{5}{7}}$$

Multiply \frac{2}{5} times \frac{5}{7} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1}{2}x-x^{2}=\frac{\frac{8}{35}}{\frac{2\times 5}{5\times 7}}$$

Cancel out 5 in both numerator and denominator.

$$\frac{1}{2}x-x^{2}=\frac{\frac{8}{35}}{\frac{2}{7}}$$

Divide \frac{8}{35} by \frac{2}{7} by multiplying \frac{8}{35} by the reciprocal of \frac{2}{7}.

$$\frac{1}{2}x-x^{2}=\frac{8}{35}\times \frac{7}{2}$$

Multiply \frac{8}{35} times \frac{7}{2} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1}{2}x-x^{2}=\frac{8\times 7}{35\times 2}$$

Do the multiplications in the fraction \frac{8\times 7}{35\times 2}.

$$\frac{1}{2}x-x^{2}=\frac{56}{70}$$

Reduce the fraction \frac{56}{70} to lowest terms by extracting and canceling out 14.

$$\frac{1}{2}x-x^{2}=\frac{4}{5}$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-x^{2}+\frac{1}{2}x=\frac{4}{5}$$

Divide both sides by -1.

$$\frac{-x^{2}+\frac{1}{2}x}{-1}=\frac{\frac{4}{5}}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{\frac{1}{2}}{-1}x=\frac{\frac{4}{5}}{-1}$$

Divide \frac{1}{2} by -1.

$$x^{2}-\frac{1}{2}x=\frac{\frac{4}{5}}{-1}$$

Divide \frac{4}{5} by -1.

$$x^{2}-\frac{1}{2}x=-\frac{4}{5}$$

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=-\frac{4}{5}+\left(-\frac{1}{4}\right)^{2}$$

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-\frac{4}{5}+\frac{1}{16}$$

Add -\frac{4}{5} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-\frac{59}{80}$$

Factor x^{2}-\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{4}\right)^{2}=-\frac{59}{80}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{-\frac{59}{80}}$$

Simplify.

$$x-\frac{1}{4}=\frac{\sqrt{295}i}{20} x-\frac{1}{4}=-\frac{\sqrt{295}i}{20}$$

Add \frac{1}{4} to both sides of the equation.

$$x=\frac{\sqrt{295}i}{20}+\frac{1}{4} x=-\frac{\sqrt{295}i}{20}+\frac{1}{4}$$