Solve (3/2t-9/2)^2+(t-3)^2/13=9/4

$$\frac { ( \frac { 3 } { 2 } t - \frac { 9 } { 2 } ) ^ { 2 } + ( t - 3 ) ^ { 2 } } { 13 } = \frac { 9 } { 4 }$$

Solve for t

$$t=6$$

$$t=0$$

Steps Using Factoring

Multiply both sides by 13.

$$\left(\frac{3}{2}t-\frac{9}{2}\right)^{2}+\left(t-3\right)^{2}=\frac{9}{4}\times 13$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{3}{2}t-\frac{9}{2}\right)^{2}.

$$\frac{9}{4}t^{2}-\frac{27}{2}t+\frac{81}{4}+\left(t-3\right)^{2}=\frac{9}{4}\times 13$$

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

$$\frac{9}{4}t^{2}-\frac{27}{2}t+\frac{81}{4}+t^{2}-6t+9=\frac{9}{4}\times 13$$

Combine \frac{9}{4}t^{2} and t^{2} to get \frac{13}{4}t^{2}.

$$\frac{13}{4}t^{2}-\frac{27}{2}t+\frac{81}{4}-6t+9=\frac{9}{4}\times 13$$

Combine -\frac{27}{2}t and -6t to get -\frac{39}{2}t.

$$\frac{13}{4}t^{2}-\frac{39}{2}t+\frac{81}{4}+9=\frac{9}{4}\times 13$$

Add \frac{81}{4} and 9 to get \frac{117}{4}.

$$\frac{13}{4}t^{2}-\frac{39}{2}t+\frac{117}{4}=\frac{9}{4}\times 13$$

Multiply \frac{9}{4} and 13 to get \frac{117}{4}.

$$\frac{13}{4}t^{2}-\frac{39}{2}t+\frac{117}{4}=\frac{117}{4}$$

Subtract \frac{117}{4} from both sides.

$$\frac{13}{4}t^{2}-\frac{39}{2}t+\frac{117}{4}-\frac{117}{4}=0$$

Subtract \frac{117}{4} from \frac{117}{4} to get 0.

$$\frac{13}{4}t^{2}-\frac{39}{2}t=0$$

Factor out t.

$$t\left(\frac{13}{4}t-\frac{39}{2}\right)=0$$

To find equation solutions, solve t=0 and \frac{13t}{4}-\frac{39}{2}=0.

$$t=0 t=6$$

Steps Using the Quadratic Formula

Multiply both sides by 13.

$$\left(\frac{3}{2}t-\frac{9}{2}\right)^{2}+\left(t-3\right)^{2}=\frac{9}{4}\times 13$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{3}{2}t-\frac{9}{2}\right)^{2}.

$$\frac{9}{4}t^{2}-\frac{27}{2}t+\frac{81}{4}+\left(t-3\right)^{2}=\frac{9}{4}\times 13$$

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

$$\frac{9}{4}t^{2}-\frac{27}{2}t+\frac{81}{4}+t^{2}-6t+9=\frac{9}{4}\times 13$$

Combine \frac{9}{4}t^{2} and t^{2} to get \frac{13}{4}t^{2}.

$$\frac{13}{4}t^{2}-\frac{27}{2}t+\frac{81}{4}-6t+9=\frac{9}{4}\times 13$$

Combine -\frac{27}{2}t and -6t to get -\frac{39}{2}t.

$$\frac{13}{4}t^{2}-\frac{39}{2}t+\frac{81}{4}+9=\frac{9}{4}\times 13$$

Add \frac{81}{4} and 9 to get \frac{117}{4}.

$$\frac{13}{4}t^{2}-\frac{39}{2}t+\frac{117}{4}=\frac{9}{4}\times 13$$

Multiply \frac{9}{4} and 13 to get \frac{117}{4}.

$$\frac{13}{4}t^{2}-\frac{39}{2}t+\frac{117}{4}=\frac{117}{4}$$

Subtract \frac{117}{4} from both sides.

$$\frac{13}{4}t^{2}-\frac{39}{2}t+\frac{117}{4}-\frac{117}{4}=0$$

Subtract \frac{117}{4} from \frac{117}{4} to get 0.

$$\frac{13}{4}t^{2}-\frac{39}{2}t=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{13}{4} for a, -\frac{39}{2} for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-\frac{39}{2}\right)±\sqrt{\left(-\frac{39}{2}\right)^{2}}}{2\times \frac{13}{4}}$$

Take the square root of \left(-\frac{39}{2}\right)^{2}.

$$t=\frac{-\left(-\frac{39}{2}\right)±\frac{39}{2}}{2\times \frac{13}{4}}$$

The opposite of -\frac{39}{2} is \frac{39}{2}.

$$t=\frac{\frac{39}{2}±\frac{39}{2}}{2\times \frac{13}{4}}$$

Multiply 2 times \frac{13}{4}.

$$t=\frac{\frac{39}{2}±\frac{39}{2}}{\frac{13}{2}}$$

Now solve the equation t=\frac{\frac{39}{2}±\frac{39}{2}}{\frac{13}{2}} when ± is plus. Add \frac{39}{2} to \frac{39}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t=\frac{39}{\frac{13}{2}}$$

Divide 39 by \frac{13}{2} by multiplying 39 by the reciprocal of \frac{13}{2}.

$$t=6$$

Now solve the equation t=\frac{\frac{39}{2}±\frac{39}{2}}{\frac{13}{2}} when ± is minus. Subtract \frac{39}{2} from \frac{39}{2} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$t=\frac{0}{\frac{13}{2}}$$

Divide 0 by \frac{13}{2} by multiplying 0 by the reciprocal of \frac{13}{2}.

$$t=0$$

The equation is now solved.

$$t=6 t=0$$

Steps for Completing the Square

Multiply both sides by 13.

$$\left(\frac{3}{2}t-\frac{9}{2}\right)^{2}+\left(t-3\right)^{2}=\frac{9}{4}\times 13$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{3}{2}t-\frac{9}{2}\right)^{2}.

$$\frac{9}{4}t^{2}-\frac{27}{2}t+\frac{81}{4}+\left(t-3\right)^{2}=\frac{9}{4}\times 13$$

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

$$\frac{9}{4}t^{2}-\frac{27}{2}t+\frac{81}{4}+t^{2}-6t+9=\frac{9}{4}\times 13$$

Combine \frac{9}{4}t^{2} and t^{2} to get \frac{13}{4}t^{2}.

$$\frac{13}{4}t^{2}-\frac{27}{2}t+\frac{81}{4}-6t+9=\frac{9}{4}\times 13$$

Combine -\frac{27}{2}t and -6t to get -\frac{39}{2}t.

$$\frac{13}{4}t^{2}-\frac{39}{2}t+\frac{81}{4}+9=\frac{9}{4}\times 13$$

Add \frac{81}{4} and 9 to get \frac{117}{4}.

$$\frac{13}{4}t^{2}-\frac{39}{2}t+\frac{117}{4}=\frac{9}{4}\times 13$$

Multiply \frac{9}{4} and 13 to get \frac{117}{4}.

$$\frac{13}{4}t^{2}-\frac{39}{2}t+\frac{117}{4}=\frac{117}{4}$$

Subtract \frac{117}{4} from both sides.

$$\frac{13}{4}t^{2}-\frac{39}{2}t=\frac{117}{4}-\frac{117}{4}$$

Subtract \frac{117}{4} from \frac{117}{4} to get 0.

$$\frac{13}{4}t^{2}-\frac{39}{2}t=0$$

Divide both sides of the equation by \frac{13}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{13}{4}t^{2}-\frac{39}{2}t}{\frac{13}{4}}=\frac{0}{\frac{13}{4}}$$

Dividing by \frac{13}{4} undoes the multiplication by \frac{13}{4}.

$$t^{2}+\left(-\frac{\frac{39}{2}}{\frac{13}{4}}\right)t=\frac{0}{\frac{13}{4}}$$

Divide -\frac{39}{2} by \frac{13}{4} by multiplying -\frac{39}{2} by the reciprocal of \frac{13}{4}.

$$t^{2}-6t=\frac{0}{\frac{13}{4}}$$

Divide 0 by \frac{13}{4} by multiplying 0 by the reciprocal of \frac{13}{4}.

$$t^{2}-6t=0$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -3.

$$t^{2}-6t+9=9$$

Factor t^{2}-6t+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-3\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-3\right)^{2}}=\sqrt{9}$$

Simplify.

$$t-3=3 t-3=-3$$

Add 3 to both sides of the equation.

$$t=6 t=0$$