Solve 2=5*t-2.25*1/2*t^2

$$2 = 5 \times t - 2.25 \times \frac { 1 } { 2 } \times t ^ { 2 }$$

Solve for t

$$t=\frac{4}{9}\approx 0.444444444$$

$$t=4$$

Steps Using the Quadratic Formula

Multiply 2.25 and \frac{1}{2} to get \frac{9}{8}.

$$2=5t-\frac{9}{8}t^{2}$$

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

$$5t-\frac{9}{8}t^{2}=2$$

Subtract 2 from both sides.

$$5t-\frac{9}{8}t^{2}-2=0$$

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.

$$-\frac{9}{8}t^{2}+5t-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{9}{8} for a, 5 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-5±\sqrt{5^{2}-4\left(-\frac{9}{8}\right)\left(-2\right)}}{2\left(-\frac{9}{8}\right)}$$

Square 5.

$$t=\frac{-5±\sqrt{25-4\left(-\frac{9}{8}\right)\left(-2\right)}}{2\left(-\frac{9}{8}\right)}$$

Multiply -4 times -\frac{9}{8}.

$$t=\frac{-5±\sqrt{25+\frac{9}{2}\left(-2\right)}}{2\left(-\frac{9}{8}\right)}$$

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

$$t=\frac{-5±\sqrt{25-9}}{2\left(-\frac{9}{8}\right)}$$

Add 25 to -9.

$$t=\frac{-5±\sqrt{16}}{2\left(-\frac{9}{8}\right)}$$

Take the square root of 16.

$$t=\frac{-5±4}{2\left(-\frac{9}{8}\right)}$$

Multiply 2 times -\frac{9}{8}.

$$t=\frac{-5±4}{-\frac{9}{4}}$$

Now solve the equation t=\frac{-5±4}{-\frac{9}{4}} when ± is plus. Add -5 to 4.

$$t=-\frac{1}{-\frac{9}{4}}$$

Divide -1 by -\frac{9}{4} by multiplying -1 by the reciprocal of -\frac{9}{4}.

$$t=\frac{4}{9}$$

Now solve the equation t=\frac{-5±4}{-\frac{9}{4}} when ± is minus. Subtract 4 from -5.

$$t=-\frac{9}{-\frac{9}{4}}$$

Divide -9 by -\frac{9}{4} by multiplying -9 by the reciprocal of -\frac{9}{4}.

$$t=4$$

The equation is now solved.

$$t=\frac{4}{9} t=4$$

Steps for Completing the Square

Multiply 2.25 and \frac{1}{2} to get \frac{9}{8}.

$$2=5t-\frac{9}{8}t^{2}$$

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

$$5t-\frac{9}{8}t^{2}=2$$

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.

$$-\frac{9}{8}t^{2}+5t=2$$

Divide both sides of the equation by -\frac{9}{8}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{9}{8}t^{2}+5t}{-\frac{9}{8}}=\frac{2}{-\frac{9}{8}}$$

Dividing by -\frac{9}{8} undoes the multiplication by -\frac{9}{8}.

$$t^{2}+\frac{5}{-\frac{9}{8}}t=\frac{2}{-\frac{9}{8}}$$

Divide 5 by -\frac{9}{8} by multiplying 5 by the reciprocal of -\frac{9}{8}.

$$t^{2}-\frac{40}{9}t=\frac{2}{-\frac{9}{8}}$$

Divide 2 by -\frac{9}{8} by multiplying 2 by the reciprocal of -\frac{9}{8}.

$$t^{2}-\frac{40}{9}t=-\frac{16}{9}$$

Divide -\frac{40}{9}, the coefficient of the x term, by 2 to get -\frac{20}{9}. Then add the square of -\frac{20}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{40}{9}t+\left(-\frac{20}{9}\right)^{2}=-\frac{16}{9}+\left(-\frac{20}{9}\right)^{2}$$

Square -\frac{20}{9} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{40}{9}t+\frac{400}{81}=-\frac{16}{9}+\frac{400}{81}$$

Add -\frac{16}{9} to \frac{400}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}-\frac{40}{9}t+\frac{400}{81}=\frac{256}{81}$$

Factor t^{2}-\frac{40}{9}t+\frac{400}{81}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{20}{9}\right)^{2}=\frac{256}{81}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{20}{9}\right)^{2}}=\sqrt{\frac{256}{81}}$$

Simplify.

$$t-\frac{20}{9}=\frac{16}{9} t-\frac{20}{9}=-\frac{16}{9}$$

Add \frac{20}{9} to both sides of the equation.

$$t=4 t=\frac{4}{9}$$