Finite Math Examples

${(t + 3)}^{2} - 12 (t + 3) + 35 = 0$

Step 1

Simplify the left side.

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Step 1.1

Simplify ${(t + 3)}^{2} - 12 (t + 3) + 35$ .

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Step 1.1.1

Simplify each term.

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Step 1.1.1.1

Apply the distributive property.

${(t + 3)}^{2} - 12 t - 12 \cdot 3 + 35 = 0$

Step 1.1.1.2

Multiply $- 12$ by $3$ .

${(t + 3)}^{2} - 12 t - 36 + 35 = 0$

Step 1.1.2

Add $- 36$ and $35$ .

${(t + 3)}^{2} - 12 t - 1 = 0$

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 $a = 1$ , $b = - 6$ , and $c = 8$ into the quadratic formula and solve for $t$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (1 \cdot 8)}}{2 \cdot 1}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Raise $- 6$ to the power of $2$ .

$t = \frac{6 \pm \sqrt{36 - 4 \cdot 1 \cdot 8}}{2 \cdot 1}$

Step 4.1.2

Multiply $- 4 \cdot 1 \cdot 8$ .

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Step 4.1.2.1

Multiply $- 4$ by $1$ .

$t = \frac{6 \pm \sqrt{36 - 4 \cdot 8}}{2 \cdot 1}$

Step 4.1.2.2

Multiply $- 4$ by $8$ .

$t = \frac{6 \pm \sqrt{36 - 32}}{2 \cdot 1}$

Step 4.1.3

Subtract $32$ from $36$ .

$t = \frac{6 \pm \sqrt{4}}{2 \cdot 1}$

Step 4.1.4

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

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

Step 4.1.5

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

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

Step 4.2

Multiply $2$ by $1$ .

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

Step 4.3

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

$t = 3 \pm 1$

Step 5

The final answer is the combination of both solutions.

$t = 4, 2$