Solve l(t)=-5t^2+80t

$$l ( t ) = - 5 t ^ { 2 } + 80 t$$

Factor

$$5t\left(16-t\right)$$

Solution Steps

Factor out 5.

$$5\left(-t^{2}+16t\right)$$

Consider -t^{2}+16t. Factor out t.

$$t\left(-t+16\right)$$

Rewrite the complete factored expression.

$$5t\left(-t+16\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$-5t^{2}+80t=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$t=\frac{-80±\sqrt{80^{2}}}{2\left(-5\right)}$$

Take the square root of 80^{2}.

$$t=\frac{-80±80}{2\left(-5\right)}$$

Multiply 2 times -5.

$$t=\frac{-80±80}{-10}$$

Now solve the equation t=\frac{-80±80}{-10} when ± is plus. Add -80 to 80.

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

Divide 0 by -10.

$$t=0$$

Now solve the equation t=\frac{-80±80}{-10} when ± is minus. Subtract 80 from -80.

$$t=-\frac{160}{-10}$$

Divide -160 by -10.

$$t=16$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 0 for x_{1} and 16 for x_{2}.

$$-5t^{2}+80t=-5t\left(t-16\right)$$

Evaluate

$$5t\left(16-t\right)$$