Finite Math Examples

$t^{2} + 14 t + 40$

Step 1

Set $t^{2} + 14 t + 40$ equal to $0$ .

$t^{2} + 14 t + 40 = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 1$ , $b = 14$ , and $c = 40$ into the quadratic formula and solve for $t$ .

$\frac{- 14 \pm \sqrt{14^{2} - 4 \cdot (1 \cdot 40)}}{2 \cdot 1}$

Step 4

Simplify.

Tap for more steps...

Step 4.1

Simplify the numerator.

Tap for more steps...

Step 4.1.1

Raise $14$ to the power of $2$ .

$t = \frac{- 14 \pm \sqrt{196 - 4 \cdot 1 \cdot 40}}{2 \cdot 1}$

Step 4.1.2

Multiply $- 4 \cdot 1 \cdot 40$ .

Tap for more steps...

Step 4.1.2.1

Multiply $- 4$ by $1$ .

$t = \frac{- 14 \pm \sqrt{196 - 4 \cdot 40}}{2 \cdot 1}$

Step 4.1.2.2

Multiply $- 4$ by $40$ .

$t = \frac{- 14 \pm \sqrt{196 - 160}}{2 \cdot 1}$

Step 4.1.3

Subtract $160$ from $196$ .

$t = \frac{- 14 \pm \sqrt{36}}{2 \cdot 1}$

Step 4.1.4

Rewrite $36$ as $6^{2}$ .

$t = \frac{- 14 \pm \sqrt{6^{2}}}{2 \cdot 1}$

Step 4.1.5

Pull terms out from under the radical, assuming positive real numbers.

$t = \frac{- 14 \pm 6}{2 \cdot 1}$

Step 4.2

Multiply $2$ by $1$ .

$t = \frac{- 14 \pm 6}{2}$

Step 4.3

Simplify $\frac{- 14 \pm 6}{2}$ .

$t = - 7 \pm 3$

Step 5

The final answer is the combination of both solutions.

$t = - 4, - 10$