Solve f(t)=28t-4t^2

$$f ( t ) = 28 t - 4 t ^ { 2 }$$

Factor

$$4t\left(7-t\right)$$

Solution Steps

Factor out 4.

$$4\left(7t-t^{2}\right)$$

Consider 7t-t^{2}. Factor out t.

$$t\left(7-t\right)$$

Rewrite the complete factored expression.

$$4t\left(-t+7\right)$$

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.

$$-4t^{2}+28t=0$$

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.

$$t=\frac{-28±\sqrt{28^{2}}}{2\left(-4\right)}$$

Take the square root of 28^{2}.

$$t=\frac{-28±28}{2\left(-4\right)}$$

Multiply 2 times -4.

$$t=\frac{-28±28}{-8}$$

Now solve the equation t=\frac{-28±28}{-8} when ± is plus. Add -28 to 28.

$$t=\frac{0}{-8}$$

Divide 0 by -8.

$$t=0$$

Now solve the equation t=\frac{-28±28}{-8} when ± is minus. Subtract 28 from -28.

$$t=-\frac{56}{-8}$$

Divide -56 by -8.

$$t=7$$

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 7 for x_{2}.

$$-4t^{2}+28t=-4t\left(t-7\right)$$

Evaluate

$$4t\left(7-t\right)$$