Solve 29/10=1/2{left(x-6/5right)}^2+32/25

$$\frac{ 29 }{ 10 } = \frac{ 1 }{ 2 } { \left(x- \frac{ 6 }{ 5 } \right) }^{ 2 } + \frac{ 32 }{ 25 }$$

Solve for x

$$x=3$$

$$x=-\frac{3}{5}=-0.6$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-\frac{6}{5}\right)^{2}.

$$\frac{29}{10}=\frac{1}{2}\left(x^{2}-\frac{12}{5}x+\frac{36}{25}\right)+\frac{32}{25}$$

Use the distributive property to multiply \frac{1}{2} by x^{2}-\frac{12}{5}x+\frac{36}{25}.

$$\frac{29}{10}=\frac{1}{2}x^{2}-\frac{6}{5}x+\frac{18}{25}+\frac{32}{25}$$

Add \frac{18}{25} and \frac{32}{25} to get 2.

$$\frac{29}{10}=\frac{1}{2}x^{2}-\frac{6}{5}x+2$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{1}{2}x^{2}-\frac{6}{5}x+2=\frac{29}{10}$$

Subtract \frac{29}{10} from both sides.

$$\frac{1}{2}x^{2}-\frac{6}{5}x+2-\frac{29}{10}=0$$

Subtract \frac{29}{10} from 2 to get -\frac{9}{10}.

$$\frac{1}{2}x^{2}-\frac{6}{5}x-\frac{9}{10}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{2} for a, -\frac{6}{5} for b, and -\frac{9}{10} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{6}{5}\right)±\sqrt{\left(-\frac{6}{5}\right)^{2}-4\times \frac{1}{2}\left(-\frac{9}{10}\right)}}{2\times \frac{1}{2}}$$

Square -\frac{6}{5} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{6}{5}\right)±\sqrt{\frac{36}{25}-4\times \frac{1}{2}\left(-\frac{9}{10}\right)}}{2\times \frac{1}{2}}$$

Multiply -4 times \frac{1}{2}.

$$x=\frac{-\left(-\frac{6}{5}\right)±\sqrt{\frac{36}{25}-2\left(-\frac{9}{10}\right)}}{2\times \frac{1}{2}}$$

Multiply -2 times -\frac{9}{10}.

$$x=\frac{-\left(-\frac{6}{5}\right)±\sqrt{\frac{36}{25}+\frac{9}{5}}}{2\times \frac{1}{2}}$$

Add \frac{36}{25} to \frac{9}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{6}{5}\right)±\sqrt{\frac{81}{25}}}{2\times \frac{1}{2}}$$

Take the square root of \frac{81}{25}.

$$x=\frac{-\left(-\frac{6}{5}\right)±\frac{9}{5}}{2\times \frac{1}{2}}$$

The opposite of -\frac{6}{5} is \frac{6}{5}.

$$x=\frac{\frac{6}{5}±\frac{9}{5}}{2\times \frac{1}{2}}$$

Multiply 2 times \frac{1}{2}.

$$x=\frac{\frac{6}{5}±\frac{9}{5}}{1}$$

Now solve the equation x=\frac{\frac{6}{5}±\frac{9}{5}}{1} when ± is plus. Add \frac{6}{5} to \frac{9}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{3}{1}$$

Divide 3 by 1.

$$x=3$$

Now solve the equation x=\frac{\frac{6}{5}±\frac{9}{5}}{1} when ± is minus. Subtract \frac{9}{5} from \frac{6}{5} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{\frac{3}{5}}{1}$$

Divide -\frac{3}{5} by 1.

$$x=-\frac{3}{5}$$

The equation is now solved.

$$x=3 x=-\frac{3}{5}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-\frac{6}{5}\right)^{2}.

$$\frac{29}{10}=\frac{1}{2}\left(x^{2}-\frac{12}{5}x+\frac{36}{25}\right)+\frac{32}{25}$$

Use the distributive property to multiply \frac{1}{2} by x^{2}-\frac{12}{5}x+\frac{36}{25}.

$$\frac{29}{10}=\frac{1}{2}x^{2}-\frac{6}{5}x+\frac{18}{25}+\frac{32}{25}$$

Add \frac{18}{25} and \frac{32}{25} to get 2.

$$\frac{29}{10}=\frac{1}{2}x^{2}-\frac{6}{5}x+2$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{1}{2}x^{2}-\frac{6}{5}x+2=\frac{29}{10}$$

Subtract 2 from both sides.

$$\frac{1}{2}x^{2}-\frac{6}{5}x=\frac{29}{10}-2$$

Subtract 2 from \frac{29}{10} to get \frac{9}{10}.

$$\frac{1}{2}x^{2}-\frac{6}{5}x=\frac{9}{10}$$

Multiply both sides by 2.

$$\frac{\frac{1}{2}x^{2}-\frac{6}{5}x}{\frac{1}{2}}=\frac{\frac{9}{10}}{\frac{1}{2}}$$

Dividing by \frac{1}{2} undoes the multiplication by \frac{1}{2}.

$$x^{2}+\left(-\frac{\frac{6}{5}}{\frac{1}{2}}\right)x=\frac{\frac{9}{10}}{\frac{1}{2}}$$

Divide -\frac{6}{5} by \frac{1}{2} by multiplying -\frac{6}{5} by the reciprocal of \frac{1}{2}.

$$x^{2}-\frac{12}{5}x=\frac{\frac{9}{10}}{\frac{1}{2}}$$

Divide \frac{9}{10} by \frac{1}{2} by multiplying \frac{9}{10} by the reciprocal of \frac{1}{2}.

$$x^{2}-\frac{12}{5}x=\frac{9}{5}$$

Divide -\frac{12}{5}, the coefficient of the x term, by 2 to get -\frac{6}{5}. Then add the square of -\frac{6}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{12}{5}x+\left(-\frac{6}{5}\right)^{2}=\frac{9}{5}+\left(-\frac{6}{5}\right)^{2}$$

Square -\frac{6}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{12}{5}x+\frac{36}{25}=\frac{9}{5}+\frac{36}{25}$$

Add \frac{9}{5} to \frac{36}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{12}{5}x+\frac{36}{25}=\frac{81}{25}$$

Factor x^{2}-\frac{12}{5}x+\frac{36}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{6}{5}\right)^{2}=\frac{81}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{6}{5}\right)^{2}}=\sqrt{\frac{81}{25}}$$

Simplify.

$$x-\frac{6}{5}=\frac{9}{5} x-\frac{6}{5}=-\frac{9}{5}$$

Add \frac{6}{5} to both sides of the equation.

$$x=3 x=-\frac{3}{5}$$