Solve s(t)=-t^2-t

$$s ( t ) = - t ^ { 2 } - t$$

Factor

$$-t\left(t+1\right)$$

Solution Steps

Factor out t.

$$t\left(-t-1\right)$$

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.

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

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.

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

Take the square root of 1.

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

The opposite of -1 is 1.

$$t=\frac{1±1}{2\left(-1\right)}$$

Multiply 2 times -1.

$$t=\frac{1±1}{-2}$$

Now solve the equation t=\frac{1±1}{-2} when ± is plus. Add 1 to 1.

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

Divide 2 by -2.

$$t=-1$$

Now solve the equation t=\frac{1±1}{-2} when ± is minus. Subtract 1 from 1.

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

Divide 0 by -2.

$$t=0$$

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

$$-t^{2}-t=-\left(t-\left(-1\right)\right)t$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$-t^{2}-t=-\left(t+1\right)t$$

Evaluate

$$-t\left(t+1\right)$$