Solve 80t-16t^2

$$80 t - 16 t ^ { 2 }$$

Factor

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

Solution Steps

Factor out 16.

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

Consider 5t-t^{2}. Factor out t.

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

Rewrite the complete factored expression.

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

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.

$$-16t^{2}+80t=0$$

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.

$$t=\frac{-80±\sqrt{80^{2}}}{2\left(-16\right)}$$

Take the square root of 80^{2}.

$$t=\frac{-80±80}{2\left(-16\right)}$$

Multiply 2 times -16.

$$t=\frac{-80±80}{-32}$$

Now solve the equation t=\frac{-80±80}{-32} when ± is plus. Add -80 to 80.

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

Divide 0 by -32.

$$t=0$$

Now solve the equation t=\frac{-80±80}{-32} when ± is minus. Subtract 80 from -80.

$$t=-\frac{160}{-32}$$

Divide -160 by -32.

$$t=5$$

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 5 for x_{2}.

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

Evaluate

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