Solve 16t^2+28t

$$16 t ^ { 2 } + 28 t$$

Factor

$$4t\left(4t+7\right)$$

Solution Steps

Factor out 4.

$$4\left(4t^{2}+7t\right)$$

Consider 4t^{2}+7t. Factor out t.

$$t\left(4t+7\right)$$

Rewrite the complete factored expression.

$$4t\left(4t+7\right)$$

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.

$$16t^{2}+28t=0$$

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.

$$t=\frac{-28±\sqrt{28^{2}}}{2\times 16}$$

Take the square root of 28^{2}.

$$t=\frac{-28±28}{2\times 16}$$

Multiply 2 times 16.

$$t=\frac{-28±28}{32}$$

Now solve the equation t=\frac{-28±28}{32} when ± is plus. Add -28 to 28.

$$t=\frac{0}{32}$$

Divide 0 by 32.

$$t=0$$

Now solve the equation t=\frac{-28±28}{32} when ± is minus. Subtract 28 from -28.

$$t=-\frac{56}{32}$$

Reduce the fraction \frac{-56}{32} to lowest terms by extracting and canceling out 8.

$$t=-\frac{7}{4}$$

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{7}{4} for x_{2}.

$$16t^{2}+28t=16t\left(t-\left(-\frac{7}{4}\right)\right)$$

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

$$16t^{2}+28t=16t\left(t+\frac{7}{4}\right)$$

Add \frac{7}{4} to t by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$16t^{2}+28t=16t\times \frac{4t+7}{4}$$

Cancel out 4, the greatest common factor in 16 and 4.

$$16t^{2}+28t=4t\left(4t+7\right)$$

Evaluate

$$4t\left(4t+7\right)$$