Solve 0=5t^2-3t

$$0 = 5 t ^ { 2 } - 3 t$$

Solve for t

$$t=\frac{3}{5}=0.6$$

$$t=0$$

Steps Using Factoring

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

$$5t^{2}-3t=0$$

Factor out t.

$$t\left(5t-3\right)=0$$

To find equation solutions, solve t=0 and 5t-3=0.

$$t=0 t=\frac{3}{5}$$

Steps Using the Quadratic Formula

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

$$5t^{2}-3t=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -3 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Take the square root of \left(-3\right)^{2}.

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

The opposite of -3 is 3.

$$t=\frac{3±3}{2\times 5}$$

Multiply 2 times 5.

$$t=\frac{3±3}{10}$$

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

$$t=\frac{6}{10}$$

Reduce the fraction \frac{6}{10} to lowest terms by extracting and canceling out 2.

$$t=\frac{3}{5}$$

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

$$t=\frac{0}{10}$$

Divide 0 by 10.

$$t=0$$

The equation is now solved.

$$t=\frac{3}{5} t=0$$

Steps for Completing the Square

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

$$5t^{2}-3t=0$$

Divide both sides by 5.

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

Dividing by 5 undoes the multiplication by 5.

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

Divide 0 by 5.

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

Divide -\frac{3}{5}, the coefficient of the x term, by 2 to get -\frac{3}{10}. Then add the square of -\frac{3}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{3}{5}t+\left(-\frac{3}{10}\right)^{2}=\left(-\frac{3}{10}\right)^{2}$$

Square -\frac{3}{10} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{3}{5}t+\frac{9}{100}=\frac{9}{100}$$

Factor t^{2}-\frac{3}{5}t+\frac{9}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{3}{10}\right)^{2}=\frac{9}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{3}{10}\right)^{2}}=\sqrt{\frac{9}{100}}$$

Simplify.

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

Add \frac{3}{10} to both sides of the equation.

$$t=\frac{3}{5} t=0$$