Solve for c 4^2+6^2=c^2

$4^{2} + 6^{2} = c^{2}$

Step 1

Rewrite the equation as $c^{2} = 4^{2} + 6^{2}$ .

$c^{2} = 4^{2} + 6^{2}$

Step 2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$c = \pm \sqrt{4^{2} + 6^{2}}$

Step 3

Simplify $\pm \sqrt{4^{2} + 6^{2}}$ .

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

$c = \pm \sqrt{16 + 6^{2}}$

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

$c = \pm \sqrt{16 + 36}$

Add $16$ and $36$ .

$c = \pm \sqrt{52}$

Rewrite $52$ as $2^{2} \cdot 13$ .

Factor $4$ out of $52$ .

$c = \pm \sqrt{4 (13)}$

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

$c = \pm \sqrt{2^{2} \cdot 13}$

Pull terms out from under the radical.

$c = \pm 2 \sqrt{13}$

Step 4

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$c = 2 \sqrt{13}$

Next, use the negative value of the $\pm$ to find the second solution.

$c = - 2 \sqrt{13}$

The complete solution is the result of both the positive and negative portions of the solution.

$c = 2 \sqrt{13}, - 2 \sqrt{13}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$c = 2 \sqrt{13}, - 2 \sqrt{13}$

Decimal Form:

$c = 7.21110255 \dots, - 7.21110255 \dots$