Solve {left(3/2-tright)}^2+{left(1/2+2t-1right)}^2={left(frac{5sqrt{2}}{6}right)}^2

$${ \left( \frac{ 3 }{ 2 } -t \right) }^{ 2 } + { \left( \frac{ 1 }{ 2 } +2t-1 \right) }^{ 2 } = { \left( \frac{ 5 \sqrt{ 2 } }{ 6 } \right) }^{ 2 }$$

Solve for t

$$t=\frac{1}{3}\approx 0.333333333$$

$$t=\frac{2}{3}\approx 0.666666667$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{3}{2}-t\right)^{2}.

$$\frac{9}{4}-3t+t^{2}+\left(\frac{1}{2}+2t-1\right)^{2}=\left(\frac{5\sqrt{2}}{6}\right)^{2}$$

Subtract 1 from \frac{1}{2} to get -\frac{1}{2}.

$$\frac{9}{4}-3t+t^{2}+\left(-\frac{1}{2}+2t\right)^{2}=\left(\frac{5\sqrt{2}}{6}\right)^{2}$$

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

$$\frac{9}{4}-3t+t^{2}+\frac{1}{4}-2t+4t^{2}=\left(\frac{5\sqrt{2}}{6}\right)^{2}$$

Add \frac{9}{4} and \frac{1}{4} to get \frac{5}{2}.

$$\frac{5}{2}-3t+t^{2}-2t+4t^{2}=\left(\frac{5\sqrt{2}}{6}\right)^{2}$$

Combine -3t and -2t to get -5t.

$$\frac{5}{2}-5t+t^{2}+4t^{2}=\left(\frac{5\sqrt{2}}{6}\right)^{2}$$

Combine t^{2} and 4t^{2} to get 5t^{2}.

$$\frac{5}{2}-5t+5t^{2}=\left(\frac{5\sqrt{2}}{6}\right)^{2}$$

To raise \frac{5\sqrt{2}}{6} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{5}{2}-5t+5t^{2}=\frac{\left(5\sqrt{2}\right)^{2}}{6^{2}}$$

Expand \left(5\sqrt{2}\right)^{2}.

$$\frac{5}{2}-5t+5t^{2}=\frac{5^{2}\left(\sqrt{2}\right)^{2}}{6^{2}}$$

Calculate 5 to the power of 2 and get 25.

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

The square of \sqrt{2} is 2.

$$\frac{5}{2}-5t+5t^{2}=\frac{25\times 2}{6^{2}}$$

Multiply 25 and 2 to get 50.

$$\frac{5}{2}-5t+5t^{2}=\frac{50}{6^{2}}$$

Calculate 6 to the power of 2 and get 36.

$$\frac{5}{2}-5t+5t^{2}=\frac{50}{36}$$

Reduce the fraction \frac{50}{36} to lowest terms by extracting and canceling out 2.

$$\frac{5}{2}-5t+5t^{2}=\frac{25}{18}$$

Subtract \frac{25}{18} from both sides.

$$\frac{5}{2}-5t+5t^{2}-\frac{25}{18}=0$$

Subtract \frac{25}{18} from \frac{5}{2} to get \frac{10}{9}.

$$\frac{10}{9}-5t+5t^{2}=0$$

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.

$$5t^{2}-5t+\frac{10}{9}=0$$

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

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

Square -5.

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

Multiply -4 times 5.

$$t=\frac{-\left(-5\right)±\sqrt{25-20\times \frac{10}{9}}}{2\times 5}$$

Multiply -20 times \frac{10}{9}.

$$t=\frac{-\left(-5\right)±\sqrt{25-\frac{200}{9}}}{2\times 5}$$

Add 25 to -\frac{200}{9}.

$$t=\frac{-\left(-5\right)±\sqrt{\frac{25}{9}}}{2\times 5}$$

Take the square root of \frac{25}{9}.

$$t=\frac{-\left(-5\right)±\frac{5}{3}}{2\times 5}$$

The opposite of -5 is 5.

$$t=\frac{5±\frac{5}{3}}{2\times 5}$$

Multiply 2 times 5.

$$t=\frac{5±\frac{5}{3}}{10}$$

Now solve the equation t=\frac{5±\frac{5}{3}}{10} when ± is plus. Add 5 to \frac{5}{3}.

$$t=\frac{\frac{20}{3}}{10}$$

Divide \frac{20}{3} by 10.

$$t=\frac{2}{3}$$

Now solve the equation t=\frac{5±\frac{5}{3}}{10} when ± is minus. Subtract \frac{5}{3} from 5.

$$t=\frac{\frac{10}{3}}{10}$$

Divide \frac{10}{3} by 10.

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

The equation is now solved.

$$t=\frac{2}{3} t=\frac{1}{3}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{3}{2}-t\right)^{2}.

$$\frac{9}{4}-3t+t^{2}+\left(\frac{1}{2}+2t-1\right)^{2}=\left(\frac{5\sqrt{2}}{6}\right)^{2}$$

Subtract 1 from \frac{1}{2} to get -\frac{1}{2}.

$$\frac{9}{4}-3t+t^{2}+\left(-\frac{1}{2}+2t\right)^{2}=\left(\frac{5\sqrt{2}}{6}\right)^{2}$$

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

$$\frac{9}{4}-3t+t^{2}+\frac{1}{4}-2t+4t^{2}=\left(\frac{5\sqrt{2}}{6}\right)^{2}$$

Add \frac{9}{4} and \frac{1}{4} to get \frac{5}{2}.

$$\frac{5}{2}-3t+t^{2}-2t+4t^{2}=\left(\frac{5\sqrt{2}}{6}\right)^{2}$$

Combine -3t and -2t to get -5t.

$$\frac{5}{2}-5t+t^{2}+4t^{2}=\left(\frac{5\sqrt{2}}{6}\right)^{2}$$

Combine t^{2} and 4t^{2} to get 5t^{2}.

$$\frac{5}{2}-5t+5t^{2}=\left(\frac{5\sqrt{2}}{6}\right)^{2}$$

To raise \frac{5\sqrt{2}}{6} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{5}{2}-5t+5t^{2}=\frac{\left(5\sqrt{2}\right)^{2}}{6^{2}}$$

Expand \left(5\sqrt{2}\right)^{2}.

$$\frac{5}{2}-5t+5t^{2}=\frac{5^{2}\left(\sqrt{2}\right)^{2}}{6^{2}}$$

Calculate 5 to the power of 2 and get 25.

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

The square of \sqrt{2} is 2.

$$\frac{5}{2}-5t+5t^{2}=\frac{25\times 2}{6^{2}}$$

Multiply 25 and 2 to get 50.

$$\frac{5}{2}-5t+5t^{2}=\frac{50}{6^{2}}$$

Calculate 6 to the power of 2 and get 36.

$$\frac{5}{2}-5t+5t^{2}=\frac{50}{36}$$

Reduce the fraction \frac{50}{36} to lowest terms by extracting and canceling out 2.

$$\frac{5}{2}-5t+5t^{2}=\frac{25}{18}$$

Subtract \frac{5}{2} from both sides.

$$-5t+5t^{2}=\frac{25}{18}-\frac{5}{2}$$

Subtract \frac{5}{2} from \frac{25}{18} to get -\frac{10}{9}.

$$-5t+5t^{2}=-\frac{10}{9}$$

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.

$$5t^{2}-5t=-\frac{10}{9}$$

Divide both sides by 5.

$$\frac{5t^{2}-5t}{5}=-\frac{\frac{10}{9}}{5}$$

Dividing by 5 undoes the multiplication by 5.

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

Divide -5 by 5.

$$t^{2}-t=-\frac{\frac{10}{9}}{5}$$

Divide -\frac{10}{9} by 5.

$$t^{2}-t=-\frac{2}{9}$$

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

$$t^{2}-t+\left(-\frac{1}{2}\right)^{2}=-\frac{2}{9}+\left(-\frac{1}{2}\right)^{2}$$

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

$$t^{2}-t+\frac{1}{4}=-\frac{2}{9}+\frac{1}{4}$$

Add -\frac{2}{9} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}-t+\frac{1}{4}=\frac{1}{36}$$

Factor t^{2}-t+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{1}{2}\right)^{2}=\frac{1}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{1}{2}\right)^{2}}=\sqrt{\frac{1}{36}}$$

Simplify.

$$t-\frac{1}{2}=\frac{1}{6} t-\frac{1}{2}=-\frac{1}{6}$$

Add \frac{1}{2} to both sides of the equation.

$$t=\frac{2}{3} t=\frac{1}{3}$$