Solve 1/2(1/2x+0.2)^2=2(x-0.6)(1/2x-1.4)-(1/2x-1.4)^2

$$\frac { 1 } { 2 } ( \frac { 1 } { 2 } x + 0.2 ) ^ { 2 } = 2 ( x - 0.6 ) ( \frac { 1 } { 2 } x - 1.4 ) - ( \frac { 1 } { 2 } x - 1.4 ) ^ { 2 }$$

Solve for x

$$x = \frac{4 \sqrt{129} + 42}{25} \approx 3.497250671$$

$$x=\frac{42-4\sqrt{129}}{25}\approx -0.137250671$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\frac{1}{2}x+0.2\right)^{2}.

$$\frac{1}{2}\left(\frac{1}{4}x^{2}+0.2x+0.04\right)=2\left(x-0.6\right)\left(\frac{1}{2}x-1.4\right)-\left(\frac{1}{2}x-1.4\right)^{2}$$

Use the distributive property to multiply \frac{1}{2} by \frac{1}{4}x^{2}+0.2x+0.04.

$$\frac{1}{8}x^{2}+\frac{1}{10}x+\frac{1}{50}=2\left(x-0.6\right)\left(\frac{1}{2}x-1.4\right)-\left(\frac{1}{2}x-1.4\right)^{2}$$

Use the distributive property to multiply 2 by x-0.6.

$$\frac{1}{8}x^{2}+\frac{1}{10}x+\frac{1}{50}=\left(2x-1.2\right)\left(\frac{1}{2}x-1.4\right)-\left(\frac{1}{2}x-1.4\right)^{2}$$

Use the distributive property to multiply 2x-1.2 by \frac{1}{2}x-1.4 and combine like terms.

$$\frac{1}{8}x^{2}+\frac{1}{10}x+\frac{1}{50}=x^{2}-\frac{17}{5}x+1.68-\left(\frac{1}{2}x-1.4\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{1}{2}x-1.4\right)^{2}.

$$\frac{1}{8}x^{2}+\frac{1}{10}x+\frac{1}{50}=x^{2}-\frac{17}{5}x+1.68-\left(\frac{1}{4}x^{2}-1.4x+1.96\right)$$

To find the opposite of \frac{1}{4}x^{2}-1.4x+1.96, find the opposite of each term.

$$\frac{1}{8}x^{2}+\frac{1}{10}x+\frac{1}{50}=x^{2}-\frac{17}{5}x+1.68-\frac{1}{4}x^{2}+1.4x-1.96$$

Combine x^{2} and -\frac{1}{4}x^{2} to get \frac{3}{4}x^{2}.

$$\frac{1}{8}x^{2}+\frac{1}{10}x+\frac{1}{50}=\frac{3}{4}x^{2}-\frac{17}{5}x+1.68+1.4x-1.96$$

Combine -\frac{17}{5}x and 1.4x to get -2x.

$$\frac{1}{8}x^{2}+\frac{1}{10}x+\frac{1}{50}=\frac{3}{4}x^{2}-2x+1.68-1.96$$

Subtract 1.96 from 1.68 to get -0.28.

$$\frac{1}{8}x^{2}+\frac{1}{10}x+\frac{1}{50}=\frac{3}{4}x^{2}-2x-0.28$$

Subtract \frac{3}{4}x^{2} from both sides.

$$\frac{1}{8}x^{2}+\frac{1}{10}x+\frac{1}{50}-\frac{3}{4}x^{2}=-2x-0.28$$

Combine \frac{1}{8}x^{2} and -\frac{3}{4}x^{2} to get -\frac{5}{8}x^{2}.

$$-\frac{5}{8}x^{2}+\frac{1}{10}x+\frac{1}{50}=-2x-0.28$$

Add 2x to both sides.

$$-\frac{5}{8}x^{2}+\frac{1}{10}x+\frac{1}{50}+2x=-0.28$$

Combine \frac{1}{10}x and 2x to get \frac{21}{10}x.

$$-\frac{5}{8}x^{2}+\frac{21}{10}x+\frac{1}{50}=-0.28$$

Add 0.28 to both sides.

$$-\frac{5}{8}x^{2}+\frac{21}{10}x+\frac{1}{50}+0.28=0$$

Add \frac{1}{50} and 0.28 to get \frac{3}{10}.

$$-\frac{5}{8}x^{2}+\frac{21}{10}x+\frac{3}{10}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{5}{8} for a, \frac{21}{10} for b, and \frac{3}{10} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{21}{10}±\sqrt{\left(\frac{21}{10}\right)^{2}-4\left(-\frac{5}{8}\right)\times \frac{3}{10}}}{2\left(-\frac{5}{8}\right)}$$

Square \frac{21}{10} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{21}{10}±\sqrt{\frac{441}{100}-4\left(-\frac{5}{8}\right)\times \frac{3}{10}}}{2\left(-\frac{5}{8}\right)}$$

Multiply -4 times -\frac{5}{8}.

$$x=\frac{-\frac{21}{10}±\sqrt{\frac{441}{100}+\frac{5}{2}\times \frac{3}{10}}}{2\left(-\frac{5}{8}\right)}$$

Multiply \frac{5}{2} times \frac{3}{10} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{21}{10}±\sqrt{\frac{441}{100}+\frac{3}{4}}}{2\left(-\frac{5}{8}\right)}$$

Add \frac{441}{100} to \frac{3}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{21}{10}±\sqrt{\frac{129}{25}}}{2\left(-\frac{5}{8}\right)}$$

Take the square root of \frac{129}{25}.

$$x=\frac{-\frac{21}{10}±\frac{\sqrt{129}}{5}}{2\left(-\frac{5}{8}\right)}$$

Multiply 2 times -\frac{5}{8}.

$$x=\frac{-\frac{21}{10}±\frac{\sqrt{129}}{5}}{-\frac{5}{4}}$$

Now solve the equation x=\frac{-\frac{21}{10}±\frac{\sqrt{129}}{5}}{-\frac{5}{4}} when ± is plus. Add -\frac{21}{10} to \frac{\sqrt{129}}{5}.

$$x=\frac{\frac{\sqrt{129}}{5}-\frac{21}{10}}{-\frac{5}{4}}$$

Divide -\frac{21}{10}+\frac{\sqrt{129}}{5} by -\frac{5}{4} by multiplying -\frac{21}{10}+\frac{\sqrt{129}}{5} by the reciprocal of -\frac{5}{4}.

$$x=\frac{42-4\sqrt{129}}{25}$$

Now solve the equation x=\frac{-\frac{21}{10}±\frac{\sqrt{129}}{5}}{-\frac{5}{4}} when ± is minus. Subtract \frac{\sqrt{129}}{5} from -\frac{21}{10}.

$$x=\frac{-\frac{\sqrt{129}}{5}-\frac{21}{10}}{-\frac{5}{4}}$$

Divide -\frac{21}{10}-\frac{\sqrt{129}}{5} by -\frac{5}{4} by multiplying -\frac{21}{10}-\frac{\sqrt{129}}{5} by the reciprocal of -\frac{5}{4}.

$$x=\frac{4\sqrt{129}+42}{25}$$

The equation is now solved.

$$x=\frac{42-4\sqrt{129}}{25} x=\frac{4\sqrt{129}+42}{25}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\frac{1}{2}x+0.2\right)^{2}.

$$\frac{1}{2}\left(\frac{1}{4}x^{2}+0.2x+0.04\right)=2\left(x-0.6\right)\left(\frac{1}{2}x-1.4\right)-\left(\frac{1}{2}x-1.4\right)^{2}$$

Use the distributive property to multiply \frac{1}{2} by \frac{1}{4}x^{2}+0.2x+0.04.

$$\frac{1}{8}x^{2}+\frac{1}{10}x+\frac{1}{50}=2\left(x-0.6\right)\left(\frac{1}{2}x-1.4\right)-\left(\frac{1}{2}x-1.4\right)^{2}$$

Use the distributive property to multiply 2 by x-0.6.

$$\frac{1}{8}x^{2}+\frac{1}{10}x+\frac{1}{50}=\left(2x-1.2\right)\left(\frac{1}{2}x-1.4\right)-\left(\frac{1}{2}x-1.4\right)^{2}$$

Use the distributive property to multiply 2x-1.2 by \frac{1}{2}x-1.4 and combine like terms.

$$\frac{1}{8}x^{2}+\frac{1}{10}x+\frac{1}{50}=x^{2}-\frac{17}{5}x+1.68-\left(\frac{1}{2}x-1.4\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{1}{2}x-1.4\right)^{2}.

$$\frac{1}{8}x^{2}+\frac{1}{10}x+\frac{1}{50}=x^{2}-\frac{17}{5}x+1.68-\left(\frac{1}{4}x^{2}-1.4x+1.96\right)$$

To find the opposite of \frac{1}{4}x^{2}-1.4x+1.96, find the opposite of each term.

$$\frac{1}{8}x^{2}+\frac{1}{10}x+\frac{1}{50}=x^{2}-\frac{17}{5}x+1.68-\frac{1}{4}x^{2}+1.4x-1.96$$

Combine x^{2} and -\frac{1}{4}x^{2} to get \frac{3}{4}x^{2}.

$$\frac{1}{8}x^{2}+\frac{1}{10}x+\frac{1}{50}=\frac{3}{4}x^{2}-\frac{17}{5}x+1.68+1.4x-1.96$$

Combine -\frac{17}{5}x and 1.4x to get -2x.

$$\frac{1}{8}x^{2}+\frac{1}{10}x+\frac{1}{50}=\frac{3}{4}x^{2}-2x+1.68-1.96$$

Subtract 1.96 from 1.68 to get -0.28.

$$\frac{1}{8}x^{2}+\frac{1}{10}x+\frac{1}{50}=\frac{3}{4}x^{2}-2x-0.28$$

Subtract \frac{3}{4}x^{2} from both sides.

$$\frac{1}{8}x^{2}+\frac{1}{10}x+\frac{1}{50}-\frac{3}{4}x^{2}=-2x-0.28$$

Combine \frac{1}{8}x^{2} and -\frac{3}{4}x^{2} to get -\frac{5}{8}x^{2}.

$$-\frac{5}{8}x^{2}+\frac{1}{10}x+\frac{1}{50}=-2x-0.28$$

Add 2x to both sides.

$$-\frac{5}{8}x^{2}+\frac{1}{10}x+\frac{1}{50}+2x=-0.28$$

Combine \frac{1}{10}x and 2x to get \frac{21}{10}x.

$$-\frac{5}{8}x^{2}+\frac{21}{10}x+\frac{1}{50}=-0.28$$

Subtract \frac{1}{50} from both sides.

$$-\frac{5}{8}x^{2}+\frac{21}{10}x=-0.28-\frac{1}{50}$$

Subtract \frac{1}{50} from -0.28 to get -\frac{3}{10}.

$$-\frac{5}{8}x^{2}+\frac{21}{10}x=-\frac{3}{10}$$

Divide both sides of the equation by -\frac{5}{8}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{5}{8}x^{2}+\frac{21}{10}x}{-\frac{5}{8}}=-\frac{\frac{3}{10}}{-\frac{5}{8}}$$

Dividing by -\frac{5}{8} undoes the multiplication by -\frac{5}{8}.

$$x^{2}+\frac{\frac{21}{10}}{-\frac{5}{8}}x=-\frac{\frac{3}{10}}{-\frac{5}{8}}$$

Divide \frac{21}{10} by -\frac{5}{8} by multiplying \frac{21}{10} by the reciprocal of -\frac{5}{8}.

$$x^{2}-\frac{84}{25}x=-\frac{\frac{3}{10}}{-\frac{5}{8}}$$

Divide -\frac{3}{10} by -\frac{5}{8} by multiplying -\frac{3}{10} by the reciprocal of -\frac{5}{8}.

$$x^{2}-\frac{84}{25}x=\frac{12}{25}$$

Divide -\frac{84}{25}, the coefficient of the x term, by 2 to get -\frac{42}{25}. Then add the square of -\frac{42}{25} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{84}{25}x+\left(-\frac{42}{25}\right)^{2}=\frac{12}{25}+\left(-\frac{42}{25}\right)^{2}$$

Square -\frac{42}{25} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{84}{25}x+\frac{1764}{625}=\frac{12}{25}+\frac{1764}{625}$$

Add \frac{12}{25} to \frac{1764}{625} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{84}{25}x+\frac{1764}{625}=\frac{2064}{625}$$

Factor x^{2}-\frac{84}{25}x+\frac{1764}{625}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{42}{25}\right)^{2}=\frac{2064}{625}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{42}{25}\right)^{2}}=\sqrt{\frac{2064}{625}}$$

Simplify.

$$x-\frac{42}{25}=\frac{4\sqrt{129}}{25} x-\frac{42}{25}=-\frac{4\sqrt{129}}{25}$$

Add \frac{42}{25} to both sides of the equation.

$$x=\frac{4\sqrt{129}+42}{25} x=\frac{42-4\sqrt{129}}{25}$$