Solve 2t^2-2t

$$2 t ^ { 2 } - 2 t$$

Factor

$$2t\left(t-1\right)$$

Solution Steps

Factor out 2.

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

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

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

Rewrite the complete factored expression.

$$2t\left(t-1\right)$$

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.

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

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.

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

Take the square root of \left(-2\right)^{2}.

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

The opposite of -2 is 2.

$$t=\frac{2±2}{2\times 2}$$

Multiply 2 times 2.

$$t=\frac{2±2}{4}$$

Now solve the equation t=\frac{2±2}{4} when ± is plus. Add 2 to 2.

$$t=\frac{4}{4}$$

Divide 4 by 4.

$$t=1$$

Now solve the equation t=\frac{2±2}{4} when ± is minus. Subtract 2 from 2.

$$t=\frac{0}{4}$$

Divide 0 by 4.

$$t=0$$

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}.

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

Evaluate

$$2t\left(t-1\right)$$