Solve S(t)=2t^2+3t

$$S ( t ) = 2 t ^ { 2 } + 3 t$$

Factor

$$t\left(2t+3\right)$$

Solution Steps

Factor out t.

$$t\left(2t+3\right)$$

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.

$$2t^{2}+3t=0$$

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.

$$t=\frac{-3±\sqrt{3^{2}}}{2\times 2}$$

Take the square root of 3^{2}.

$$t=\frac{-3±3}{2\times 2}$$

Multiply 2 times 2.

$$t=\frac{-3±3}{4}$$

Now solve the equation t=\frac{-3±3}{4} when ± is plus. Add -3 to 3.

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

Divide 0 by 4.

$$t=0$$

Now solve the equation t=\frac{-3±3}{4} when ± is minus. Subtract 3 from -3.

$$t=-\frac{6}{4}$$

Reduce the fraction \frac{-6}{4} to lowest terms by extracting and canceling out 2.

$$t=-\frac{3}{2}$$

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 -\frac{3}{2} for x_{2}.

$$2t^{2}+3t=2t\left(t-\left(-\frac{3}{2}\right)\right)$$

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

$$2t^{2}+3t=2t\left(t+\frac{3}{2}\right)$$

Add \frac{3}{2} to t by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$2t^{2}+3t=2t\times \frac{2t+3}{2}$$

Cancel out 2, the greatest common factor in 2 and 2.

$$2t^{2}+3t=t\left(2t+3\right)$$

Evaluate

$$t\left(2t+3\right)$$