Solve Using the Quadratic Formula a^2+24^2=26^2

$a^{2} + 24^{2} = 26^{2}$

Step 1

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

Simplify the left side.

Raise $24$ to the power of $2$ .

$a^{2} + 576 = 26^{2}$

Simplify the right side.

Raise $26$ to the power of $2$ .

$a^{2} + 576 = 676$

Subtract $676$ from both sides of the equation.

$a^{2} + 576 - 676 = 0$

Subtract $676$ from $576$ .

$a^{2} - 100 = 0$

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 $a$ .

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

Step 4

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $- 100$ .

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

Add $0$ and $400$ .

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

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

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

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

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

Multiply $2$ by $1$ .

$a = \frac{0 \pm 20}{2}$

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

$a = \pm 10$

Step 5

The final answer is the combination of both solutions.

$a = 10, - 10$