Solve Using the Quadratic Formula 6^2+8^2=c^2

$6^{2} + 8^{2} = c^{2}$

Step 1

Move all terms to the left side of the equation and simplify.

Simplify the left side.

Simplify $6^{2} + 8^{2}$ .

Simplify each term.

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

$36 + 8^{2} = c^{2}$

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

$36 + 64 = c^{2}$

Add $36$ and $64$ .

$100 = c^{2}$

Subtract $c^{2}$ from both sides of the equation.

$100 - c^{2} = 0$

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 $a = - 1$ , $b = 0$ , and $c = 100$ into the quadratic formula and solve for $c$ .

$\frac{0 \pm \sqrt{0^{2} - 4 \cdot (- 1 \cdot 100)}}{2 \cdot - 1}$

Step 4

Simplify.

Simplify the numerator.

Raising $0$ to any positive power yields $0$ .

$c = \frac{0 \pm \sqrt{0 - 4 \cdot - 1 \cdot 100}}{2 \cdot - 1}$

Multiply $- 4 \cdot - 1 \cdot 100$ .

Multiply $- 4$ by $- 1$ .

$c = \frac{0 \pm \sqrt{0 + 4 \cdot 100}}{2 \cdot - 1}$

Multiply $4$ by $100$ .

$c = \frac{0 \pm \sqrt{0 + 400}}{2 \cdot - 1}$

Add $0$ and $400$ .

$c = \frac{0 \pm \sqrt{400}}{2 \cdot - 1}$

Rewrite $400$ as $20^{2}$ .

$c = \frac{0 \pm \sqrt{20^{2}}}{2 \cdot - 1}$

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

$c = \frac{0 \pm 20}{2 \cdot - 1}$

Multiply $2$ by $- 1$ .

$c = \frac{0 \pm 20}{- 2}$

Simplify $\frac{0 \pm 20}{- 2}$ .

$c = \pm 10$

Step 5

The final answer is the combination of both solutions.

$c = 10, - 10$