Finite Math Examples

$4 g^{2} - 8 g + c = 0$

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 $a = 4$ , $b = - 8$ , and $c = c$ into the quadratic formula and solve for $g$ .

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

Step 3

Simplify.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Raise $- 8$ to the power of $2$ .

$g = \frac{8 \pm \sqrt{64 - 4 \cdot 4 \cdot c}}{2 \cdot 4}$

Step 3.1.2

Multiply $- 4$ by $4$ .

$g = \frac{8 \pm \sqrt{64 - 16 \cdot c}}{2 \cdot 4}$

Step 3.1.3

Factor $16$ out of $64 - 16 c$ .

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Step 3.1.3.1

Factor $16$ out of $64$ .

$g = \frac{8 \pm \sqrt{16 (4) - 16 c}}{2 \cdot 4}$

Step 3.1.3.2

Factor $16$ out of $- 16 c$ .

$g = \frac{8 \pm \sqrt{16 (4) + 16 (- c)}}{2 \cdot 4}$

Step 3.1.3.3

Factor $16$ out of $16 (4) + 16 (- c)$ .

$g = \frac{8 \pm \sqrt{16 (4 - c)}}{2 \cdot 4}$

Step 3.1.4

Rewrite $16 (4 - c)$ as ${(2^{2})}^{2} (4 - c)$ .

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Step 3.1.4.1

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

$g = \frac{8 \pm \sqrt{4^{2} (4 - c)}}{2 \cdot 4}$

Step 3.1.4.2

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

$g = \frac{8 \pm \sqrt{{(2^{2})}^{2} (4 - c)}}{2 \cdot 4}$

Step 3.1.5

Pull terms out from under the radical.

$g = \frac{8 \pm 2^{2} \sqrt{4 - c}}{2 \cdot 4}$

Step 3.1.6

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

$g = \frac{8 \pm 4 \sqrt{4 - c}}{2 \cdot 4}$

Step 3.2

Multiply $2$ by $4$ .

$g = \frac{8 \pm 4 \sqrt{4 - c}}{8}$

Step 3.3

Simplify $\frac{8 \pm 4 \sqrt{4 - c}}{8}$ .

$g = \frac{2 \pm \sqrt{4 - c}}{2}$

Step 4

The final answer is the combination of both solutions.

$g = \frac{2 + \sqrt{4 - c}}{2}$

$g = \frac{2 - \sqrt{4 - c}}{2}$