Solve 5t^2+10t

$$5 t ^ { 2 } + 10 t$$

Factor

$$5t\left(t+2\right)$$

Solution Steps

Factor out 5.

$$5\left(t^{2}+2t\right)$$

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

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

Rewrite the complete factored expression.

$$5t\left(t+2\right)$$

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.

$$5t^{2}+10t=0$$

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.

$$t=\frac{-10±\sqrt{10^{2}}}{2\times 5}$$

Take the square root of 10^{2}.

$$t=\frac{-10±10}{2\times 5}$$

Multiply 2 times 5.

$$t=\frac{-10±10}{10}$$

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

$$t=\frac{0}{10}$$

Divide 0 by 10.

$$t=0$$

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

$$t=-\frac{20}{10}$$

Divide -20 by 10.

$$t=-2$$

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}.

$$5t^{2}+10t=5t\left(t-\left(-2\right)\right)$$

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

$$5t^{2}+10t=5t\left(t+2\right)$$

Evaluate

$$5t\left(t+2\right)$$