Solve 0=-5t^2+40t

$$0 = - 5 t ^ { 2 } + 40 t$$

Solve for t

$$t=8$$

$$t=0$$

Steps Using Factoring

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

$$-5t^{2}+40t=0$$

Factor out t.

$$t\left(-5t+40\right)=0$$

To find equation solutions, solve t=0 and -5t+40=0.

$$t=0 t=8$$

Steps Using the Quadratic Formula

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

$$-5t^{2}+40t=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 40 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-40±\sqrt{40^{2}}}{2\left(-5\right)}$$

Take the square root of 40^{2}.

$$t=\frac{-40±40}{2\left(-5\right)}$$

Multiply 2 times -5.

$$t=\frac{-40±40}{-10}$$

Now solve the equation t=\frac{-40±40}{-10} when ± is plus. Add -40 to 40.

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

Divide 0 by -10.

$$t=0$$

Now solve the equation t=\frac{-40±40}{-10} when ± is minus. Subtract 40 from -40.

$$t=-\frac{80}{-10}$$

Divide -80 by -10.

$$t=8$$

The equation is now solved.

$$t=0 t=8$$

Steps for Completing the Square

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

$$-5t^{2}+40t=0$$

Divide both sides by -5.

$$\frac{-5t^{2}+40t}{-5}=\frac{0}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$t^{2}+\frac{40}{-5}t=\frac{0}{-5}$$

Divide 40 by -5.

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

Divide 0 by -5.

$$t^{2}-8t=0$$

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

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

Square -4.

$$t^{2}-8t+16=16$$

Factor t^{2}-8t+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$t-4=4 t-4=-4$$

Add 4 to both sides of the equation.

$$t=8 t=0$$