Solve (2-1/2t)^2+(-1+1/2t)^2=4

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

Solve for t

$$t=\sqrt{7}+3\approx 5.645751311$$

$$t=3-\sqrt{7}\approx 0.354248689$$

Steps Using the Quadratic Formula

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

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

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

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

Add 4 and 1 to get 5.

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

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

$$5-3t+\frac{1}{4}t^{2}+\frac{1}{4}t^{2}=4$$

Combine \frac{1}{4}t^{2} and \frac{1}{4}t^{2} to get \frac{1}{2}t^{2}.

$$5-3t+\frac{1}{2}t^{2}=4$$

Subtract 4 from both sides.

$$5-3t+\frac{1}{2}t^{2}-4=0$$

Subtract 4 from 5 to get 1.

$$1-3t+\frac{1}{2}t^{2}=0$$

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.

$$\frac{1}{2}t^{2}-3t+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{2} for a, -3 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times \frac{1}{2}}}{2\times \frac{1}{2}}$$

Square -3.

$$t=\frac{-\left(-3\right)±\sqrt{9-4\times \frac{1}{2}}}{2\times \frac{1}{2}}$$

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

$$t=\frac{-\left(-3\right)±\sqrt{9-2}}{2\times \frac{1}{2}}$$

Add 9 to -2.

$$t=\frac{-\left(-3\right)±\sqrt{7}}{2\times \frac{1}{2}}$$

The opposite of -3 is 3.

$$t=\frac{3±\sqrt{7}}{2\times \frac{1}{2}}$$

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

$$t=\frac{3±\sqrt{7}}{1}$$

Now solve the equation t=\frac{3±\sqrt{7}}{1} when ± is plus. Add 3 to \sqrt{7}.

$$t=\frac{\sqrt{7}+3}{1}$$

Divide 3+\sqrt{7} by 1.

$$t=\sqrt{7}+3$$

Now solve the equation t=\frac{3±\sqrt{7}}{1} when ± is minus. Subtract \sqrt{7} from 3.

$$t=\frac{3-\sqrt{7}}{1}$$

Divide 3-\sqrt{7} by 1.

$$t=3-\sqrt{7}$$

The equation is now solved.

$$t=\sqrt{7}+3 t=3-\sqrt{7}$$

Steps for Completing the Square

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

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

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

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

Add 4 and 1 to get 5.

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

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

$$5-3t+\frac{1}{4}t^{2}+\frac{1}{4}t^{2}=4$$

Combine \frac{1}{4}t^{2} and \frac{1}{4}t^{2} to get \frac{1}{2}t^{2}.

$$5-3t+\frac{1}{2}t^{2}=4$$

Subtract 5 from both sides.

$$-3t+\frac{1}{2}t^{2}=4-5$$

Subtract 5 from 4 to get -1.

$$-3t+\frac{1}{2}t^{2}=-1$$

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.

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

Multiply both sides by 2.

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

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

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

Divide -3 by \frac{1}{2} by multiplying -3 by the reciprocal of \frac{1}{2}.

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

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

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

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.

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

Square -3.

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

Add -2 to 9.

$$t^{2}-6t+9=7$$

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}.

$$\left(t-3\right)^{2}=7$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-3\right)^{2}}=\sqrt{7}$$

Simplify.

$$t-3=\sqrt{7} t-3=-\sqrt{7}$$

Add 3 to both sides of the equation.

$$t=\sqrt{7}+3 t=3-\sqrt{7}$$