Solve h(t)=-t^2+6t

$$h ( t ) = - t ^ { 2 } + 6 t$$

Factor

$$t\left(6-t\right)$$

Solution Steps

Factor out t.

$$t\left(-t+6\right)$$

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.

$$-t^{2}+6t=0$$

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.

$$t=\frac{-6±\sqrt{6^{2}}}{2\left(-1\right)}$$

Take the square root of 6^{2}.

$$t=\frac{-6±6}{2\left(-1\right)}$$

Multiply 2 times -1.

$$t=\frac{-6±6}{-2}$$

Now solve the equation t=\frac{-6±6}{-2} when ± is plus. Add -6 to 6.

$$t=\frac{0}{-2}$$

Divide 0 by -2.

$$t=0$$

Now solve the equation t=\frac{-6±6}{-2} when ± is minus. Subtract 6 from -6.

$$t=-\frac{12}{-2}$$

Divide -12 by -2.

$$t=6$$

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 6 for x_{2}.

$$-t^{2}+6t=-t\left(t-6\right)$$

Evaluate

$$t\left(6-t\right)$$