Solve Using the Quadratic Formula 5^2+b^2=13^2

$5^{2} + b^{2} = 13^{2}$

Step 1

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

Simplify the left side.

Raise $5$ to the power of $2$ .

$25 + b^{2} = 13^{2}$

Simplify the right side.

Raise $13$ to the power of $2$ .

$25 + b^{2} = 169$

Subtract $169$ from both sides of the equation.

$25 + b^{2} - 169 = 0$

Subtract $169$ from $25$ .

$b^{2} - 144 = 0$

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 $c = - 144$ into the quadratic formula and solve for $b$ .

$\frac{0 \pm \sqrt{0^{2} - 4 \cdot (1 \cdot - 144)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

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

$b = \frac{0 \pm \sqrt{0 - 4 \cdot 1 \cdot - 144}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 144$ .

Multiply $- 4$ by $1$ .

$b = \frac{0 \pm \sqrt{0 - 4 \cdot - 144}}{2 \cdot 1}$

Multiply $- 4$ by $- 144$ .

$b = \frac{0 \pm \sqrt{0 + 576}}{2 \cdot 1}$

Add $0$ and $576$ .

$b = \frac{0 \pm \sqrt{576}}{2 \cdot 1}$

Rewrite $576$ as $24^{2}$ .

$b = \frac{0 \pm \sqrt{24^{2}}}{2 \cdot 1}$

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

$b = \frac{0 \pm 24}{2 \cdot 1}$

Multiply $2$ by $1$ .

$b = \frac{0 \pm 24}{2}$

Simplify $\frac{0 \pm 24}{2}$ .

$b = \pm 12$

Step 5

The final answer is the combination of both solutions.

$b = 12, - 12$