Solve 5t^2+t

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

Factor

$$t\left(5t+1\right)$$

Solution Steps

Factor out t.

$$t\left(5t+1\right)$$

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.

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

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.

$$t=\frac{-1±\sqrt{1^{2}}}{2\times 5}$$

Take the square root of 1^{2}.

$$t=\frac{-1±1}{2\times 5}$$

Multiply 2 times 5.

$$t=\frac{-1±1}{10}$$

Now solve the equation t=\frac{-1±1}{10} when ± is plus. Add -1 to 1.

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

Divide 0 by 10.

$$t=0$$

Now solve the equation t=\frac{-1±1}{10} when ± is minus. Subtract 1 from -1.

$$t=-\frac{2}{10}$$

Reduce the fraction \frac{-2}{10} to lowest terms by extracting and canceling out 2.

$$t=-\frac{1}{5}$$

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 -\frac{1}{5} for x_{2}.

$$5t^{2}+t=5t\left(t-\left(-\frac{1}{5}\right)\right)$$

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

$$5t^{2}+t=5t\left(t+\frac{1}{5}\right)$$

Add \frac{1}{5} to t by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$5t^{2}+t=5t\times \frac{5t+1}{5}$$

Cancel out 5, the greatest common factor in 5 and 5.

$$5t^{2}+t=t\left(5t+1\right)$$

Evaluate

$$t\left(5t+1\right)$$