Solve 3t(5t-3)=0

$$3 t ( 5 t - 3 ) = 0$$

Solve for t

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

$$t=0$$

Steps Using Factoring

Use the distributive property to multiply 3t by 5t-3.

$$15t^{2}-9t=0$$

Factor out t.

$$t\left(15t-9\right)=0$$

To find equation solutions, solve t=0 and 15t-9=0.

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

Steps Using the Quadratic Formula

Use the distributive property to multiply 3t by 5t-3.

$$15t^{2}-9t=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 15 for a, -9 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}}}{2\times 15}$$

Take the square root of \left(-9\right)^{2}.

$$t=\frac{-\left(-9\right)±9}{2\times 15}$$

The opposite of -9 is 9.

$$t=\frac{9±9}{2\times 15}$$

Multiply 2 times 15.

$$t=\frac{9±9}{30}$$

Now solve the equation t=\frac{9±9}{30} when ± is plus. Add 9 to 9.

$$t=\frac{18}{30}$$

Reduce the fraction \frac{18}{30} to lowest terms by extracting and canceling out 6.

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

Now solve the equation t=\frac{9±9}{30} when ± is minus. Subtract 9 from 9.

$$t=\frac{0}{30}$$

Divide 0 by 30.

$$t=0$$

The equation is now solved.

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

Steps for Completing the Square

Use the distributive property to multiply 3t by 5t-3.

$$15t^{2}-9t=0$$

Divide both sides by 15.

$$\frac{15t^{2}-9t}{15}=\frac{0}{15}$$

Dividing by 15 undoes the multiplication by 15.

$$t^{2}+\left(-\frac{9}{15}\right)t=\frac{0}{15}$$

Reduce the fraction \frac{-9}{15} to lowest terms by extracting and canceling out 3.

$$t^{2}-\frac{3}{5}t=\frac{0}{15}$$

Divide 0 by 15.

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

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.

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