Solve Using the Quadratic Formula b(b+17)=0

$b (b + 17) = 0$

Step 1

Simplify the left side.

Simplify $b (b + 17)$ .

Apply the distributive property.

$b \cdot b + b \cdot 17 = 0$

Simplify the expression.

Multiply $b$ by $b$ .

$b^{2} + b \cdot 17 = 0$

Move $17$ to the left of $b$ .

$b^{2} + 17 b = 0$

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 $a = 1$ , $b = 17$ , and $c = 0$ into the quadratic formula and solve for $b$ .

$\frac{- 17 \pm \sqrt{17^{2} - 4 \cdot (1 \cdot 0)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $17$ to the power of $2$ .

$b = \frac{- 17 \pm \sqrt{289 - 4 \cdot 1 \cdot 0}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 0$ .

Multiply $- 4$ by $1$ .

$b = \frac{- 17 \pm \sqrt{289 - 4 \cdot 0}}{2 \cdot 1}$

Multiply $- 4$ by $0$ .

$b = \frac{- 17 \pm \sqrt{289 + 0}}{2 \cdot 1}$

Add $289$ and $0$ .

$b = \frac{- 17 \pm \sqrt{289}}{2 \cdot 1}$

Rewrite $289$ as $17^{2}$ .

$b = \frac{- 17 \pm \sqrt{17^{2}}}{2 \cdot 1}$

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

$b = \frac{- 17 \pm 17}{2 \cdot 1}$

Multiply $2$ by $1$ .

$b = \frac{- 17 \pm 17}{2}$

Step 5

The final answer is the combination of both solutions.

$b = 0, - 17$