Solve 0=9t^2-62t

$$0 = 9 t ^ { 2 } - 62 t$$

Solve for t

$$t = \frac{62}{9} = 6\frac{8}{9} \approx 6.888888889$$

$$t=0$$

Steps Using Factoring

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

$$9t^{2}-62t=0$$

Factor out t.

$$t\left(9t-62\right)=0$$

To find equation solutions, solve t=0 and 9t-62=0.

$$t=0 t=\frac{62}{9}$$

Steps Using the Quadratic Formula

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

$$9t^{2}-62t=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -62 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-62\right)±\sqrt{\left(-62\right)^{2}}}{2\times 9}$$

Take the square root of \left(-62\right)^{2}.

$$t=\frac{-\left(-62\right)±62}{2\times 9}$$

The opposite of -62 is 62.

$$t=\frac{62±62}{2\times 9}$$

Multiply 2 times 9.

$$t=\frac{62±62}{18}$$

Now solve the equation t=\frac{62±62}{18} when ± is plus. Add 62 to 62.

$$t=\frac{124}{18}$$

Reduce the fraction \frac{124}{18} to lowest terms by extracting and canceling out 2.

$$t=\frac{62}{9}$$

Now solve the equation t=\frac{62±62}{18} when ± is minus. Subtract 62 from 62.

$$t=\frac{0}{18}$$

Divide 0 by 18.

$$t=0$$

The equation is now solved.

$$t=\frac{62}{9} t=0$$

Steps for Completing the Square

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

$$9t^{2}-62t=0$$

Divide both sides by 9.

$$\frac{9t^{2}-62t}{9}=\frac{0}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$t^{2}-\frac{62}{9}t=\frac{0}{9}$$

Divide 0 by 9.

$$t^{2}-\frac{62}{9}t=0$$

Divide -\frac{62}{9}, the coefficient of the x term, by 2 to get -\frac{31}{9}. Then add the square of -\frac{31}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{62}{9}t+\left(-\frac{31}{9}\right)^{2}=\left(-\frac{31}{9}\right)^{2}$$

Square -\frac{31}{9} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{62}{9}t+\frac{961}{81}=\frac{961}{81}$$

Factor t^{2}-\frac{62}{9}t+\frac{961}{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}.

$$\left(t-\frac{31}{9}\right)^{2}=\frac{961}{81}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{31}{9}\right)^{2}}=\sqrt{\frac{961}{81}}$$

Simplify.

$$t-\frac{31}{9}=\frac{31}{9} t-\frac{31}{9}=-\frac{31}{9}$$

Add \frac{31}{9} to both sides of the equation.

$$t=\frac{62}{9} t=0$$