Solve 2t^2+4t

$$2 t ^ { 2 } + 4 t$$

Factor

$$2t\left(t+2\right)$$

Solution Steps

Factor out 2.

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

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

$$t\left(t+2\right)$$

Rewrite the complete factored expression.

$$2t\left(t+2\right)$$

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.

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

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.

$$t=\frac{-4±\sqrt{4^{2}}}{2\times 2}$$

Take the square root of 4^{2}.

$$t=\frac{-4±4}{2\times 2}$$

Multiply 2 times 2.

$$t=\frac{-4±4}{4}$$

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

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

Divide 0 by 4.

$$t=0$$

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

$$t=-\frac{8}{4}$$

Divide -8 by 4.

$$t=-2$$

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 -2 for x_{2}.

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

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

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

Evaluate

$$2t\left(t+2\right)$$