Solve -1/2t+2t-t^2+7/2t+2=2

$$- \frac { 1 } { 2 } t + 2 t - t ^ { 2 } + \frac { 7 } { 2 } t + 2 = 2$$

Solve for t

$$t=5$$

$$t=0$$

Steps Using Factoring

Combine -\frac{1}{2}t and 2t to get \frac{3}{2}t.

$$\frac{3}{2}t-t^{2}+\frac{7}{2}t+2=2$$

Combine \frac{3}{2}t and \frac{7}{2}t to get 5t.

$$5t-t^{2}+2=2$$

Subtract 2 from both sides.

$$5t-t^{2}+2-2=0$$

Subtract 2 from 2 to get 0.

$$5t-t^{2}=0$$

Factor out t.

$$t\left(5-t\right)=0$$

To find equation solutions, solve t=0 and 5-t=0.

$$t=0 t=5$$

Steps Using the Quadratic Formula

Combine -\frac{1}{2}t and 2t to get \frac{3}{2}t.

$$\frac{3}{2}t-t^{2}+\frac{7}{2}t+2=2$$

Combine \frac{3}{2}t and \frac{7}{2}t to get 5t.

$$5t-t^{2}+2=2$$

Subtract 2 from both sides.

$$5t-t^{2}+2-2=0$$

Subtract 2 from 2 to get 0.

$$5t-t^{2}=0$$

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.

$$-t^{2}+5t=0$$

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

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

Take the square root of 5^{2}.

$$t=\frac{-5±5}{2\left(-1\right)}$$

Multiply 2 times -1.

$$t=\frac{-5±5}{-2}$$

Now solve the equation t=\frac{-5±5}{-2} when ± is plus. Add -5 to 5.

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

Divide 0 by -2.

$$t=0$$

Now solve the equation t=\frac{-5±5}{-2} when ± is minus. Subtract 5 from -5.

$$t=-\frac{10}{-2}$$

Divide -10 by -2.

$$t=5$$

The equation is now solved.

$$t=0 t=5$$

Steps for Completing the Square

Combine -\frac{1}{2}t and 2t to get \frac{3}{2}t.

$$\frac{3}{2}t-t^{2}+\frac{7}{2}t+2=2$$

Combine \frac{3}{2}t and \frac{7}{2}t to get 5t.

$$5t-t^{2}+2=2$$

Subtract 2 from both sides.

$$5t-t^{2}=2-2$$

Subtract 2 from 2 to get 0.

$$5t-t^{2}=0$$

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.

$$-t^{2}+5t=0$$

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 5 by -1.

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

Divide 0 by -1.

$$t^{2}-5t=0$$

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

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

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-5t+\frac{25}{4}=\frac{25}{4}$$

Factor t^{2}-5t+\frac{25}{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}.

$$\left(t-\frac{5}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{5}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$t-\frac{5}{2}=\frac{5}{2} t-\frac{5}{2}=-\frac{5}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$t=5 t=0$$