Finite Math Examples

$12 t^{2} - 4 t + 5 = 0$

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

Substitute the values $a = 12$ , $b = - 4$ , and $c = 5$ into the quadratic formula and solve for $t$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (12 \cdot 5)}}{2 \cdot 12}$

Step 3

Simplify.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Raise $- 4$ to the power of $2$ .

$t = \frac{4 \pm \sqrt{16 - 4 \cdot 12 \cdot 5}}{2 \cdot 12}$

Step 3.1.2

Multiply $- 4 \cdot 12 \cdot 5$ .

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Step 3.1.2.1

Multiply $- 4$ by $12$ .

$t = \frac{4 \pm \sqrt{16 - 48 \cdot 5}}{2 \cdot 12}$

Step 3.1.2.2

Multiply $- 48$ by $5$ .

$t = \frac{4 \pm \sqrt{16 - 240}}{2 \cdot 12}$

Step 3.1.3

Subtract $240$ from $16$ .

$t = \frac{4 \pm \sqrt{- 224}}{2 \cdot 12}$

Step 3.1.4

Rewrite $- 224$ as $- 1 (224)$ .

$t = \frac{4 \pm \sqrt{- 1 \cdot 224}}{2 \cdot 12}$

Step 3.1.5

Rewrite $\sqrt{- 1 (224)}$ as $\sqrt{- 1} \cdot \sqrt{224}$ .

$t = \frac{4 \pm \sqrt{- 1} \cdot \sqrt{224}}{2 \cdot 12}$

Step 3.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$t = \frac{4 \pm i \cdot \sqrt{224}}{2 \cdot 12}$

Step 3.1.7

Rewrite $224$ as $4^{2} \cdot 14$ .

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Step 3.1.7.1

Factor $16$ out of $224$ .

$t = \frac{4 \pm i \cdot \sqrt{16 (14)}}{2 \cdot 12}$

Step 3.1.7.2

Rewrite $16$ as $4^{2}$ .

$t = \frac{4 \pm i \cdot \sqrt{4^{2} \cdot 14}}{2 \cdot 12}$

Step 3.1.8

Pull terms out from under the radical.

$t = \frac{4 \pm i \cdot (4 \sqrt{14})}{2 \cdot 12}$

Step 3.1.9

Move $4$ to the left of $i$ .

$t = \frac{4 \pm 4 i \sqrt{14}}{2 \cdot 12}$

Step 3.2

Multiply $2$ by $12$ .

$t = \frac{4 \pm 4 i \sqrt{14}}{24}$

Step 3.3

Simplify $\frac{4 \pm 4 i \sqrt{14}}{24}$ .

$t = \frac{1 \pm i \sqrt{14}}{6}$