Solve -t^2+10t=

$$- t ^ { 2 } + 10 t =$$

Factor

$$t\left(10-t\right)$$

Solution Steps

Factor out t.

$$t\left(-t+10\right)$$

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.

$$-t^{2}+10t=0$$

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.

$$t=\frac{-10±\sqrt{10^{2}}}{2\left(-1\right)}$$

Take the square root of 10^{2}.

$$t=\frac{-10±10}{2\left(-1\right)}$$

Multiply 2 times -1.

$$t=\frac{-10±10}{-2}$$

Now solve the equation t=\frac{-10±10}{-2} when ± is plus. Add -10 to 10.

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

Divide 0 by -2.

$$t=0$$

Now solve the equation t=\frac{-10±10}{-2} when ± is minus. Subtract 10 from -10.

$$t=-\frac{20}{-2}$$

Divide -20 by -2.

$$t=10$$

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 10 for x_{2}.

$$-t^{2}+10t=-t\left(t-10\right)$$

Evaluate

$$t\left(10-t\right)$$