Solve 0=-16t^2+4t

$$0=-16 { t }^{ 2 } +4t$$

Solve for t

$$t=\frac{1}{4}=0.25$$

$$t=0$$

Steps Using Factoring

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

$$-16t^{2}+4t=0$$

Factor out t.

$$t\left(-16t+4\right)=0$$

To find equation solutions, solve t=0 and -16t+4=0.

$$t=0 t=\frac{1}{4}$$

Steps Using the Quadratic Formula

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

$$-16t^{2}+4t=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -16 for a, 4 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-4±\sqrt{4^{2}}}{2\left(-16\right)}$$

Take the square root of 4^{2}.

$$t=\frac{-4±4}{2\left(-16\right)}$$

Multiply 2 times -16.

$$t=\frac{-4±4}{-32}$$

Now solve the equation t=\frac{-4±4}{-32} when ± is plus. Add -4 to 4.

$$t=\frac{0}{-32}$$

Divide 0 by -32.

$$t=0$$

Now solve the equation t=\frac{-4±4}{-32} when ± is minus. Subtract 4 from -4.

$$t=-\frac{8}{-32}$$

Reduce the fraction \frac{-8}{-32} to lowest terms by extracting and canceling out 8.

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

The equation is now solved.

$$t=0 t=\frac{1}{4}$$

Steps for Completing the Square

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

$$-16t^{2}+4t=0$$

Divide both sides by -16.

$$\frac{-16t^{2}+4t}{-16}=\frac{0}{-16}$$

Dividing by -16 undoes the multiplication by -16.

$$t^{2}+\frac{4}{-16}t=\frac{0}{-16}$$

Reduce the fraction \frac{4}{-16} to lowest terms by extracting and canceling out 4.

$$t^{2}-\frac{1}{4}t=\frac{0}{-16}$$

Divide 0 by -16.

$$t^{2}-\frac{1}{4}t=0$$

Divide -\frac{1}{4}, the coefficient of the x term, by 2 to get -\frac{1}{8}. Then add the square of -\frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{1}{4}t+\left(-\frac{1}{8}\right)^{2}=\left(-\frac{1}{8}\right)^{2}$$

Square -\frac{1}{8} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{1}{4}t+\frac{1}{64}=\frac{1}{64}$$

Factor t^{2}-\frac{1}{4}t+\frac{1}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{1}{8}\right)^{2}=\frac{1}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{1}{8}\right)^{2}}=\sqrt{\frac{1}{64}}$$

Simplify.

$$t-\frac{1}{8}=\frac{1}{8} t-\frac{1}{8}=-\frac{1}{8}$$

Add \frac{1}{8} to both sides of the equation.

$$t=\frac{1}{4} t=0$$