Solve Using the Quadratic Formula b^2+b-30=0

$b^{2} + b - 30 = 0$

Step 1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2

Substitute the values $a = 1$ , $b = 1$ , and $c = - 30$ into the quadratic formula and solve for $b$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (1 \cdot - 30)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

One to any power is one.

$b = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot - 30}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 30$ .

Multiply $- 4$ by $1$ .

$b = \frac{- 1 \pm \sqrt{1 - 4 \cdot - 30}}{2 \cdot 1}$

Multiply $- 4$ by $- 30$ .

$b = \frac{- 1 \pm \sqrt{1 + 120}}{2 \cdot 1}$

Add $1$ and $120$ .

$b = \frac{- 1 \pm \sqrt{121}}{2 \cdot 1}$

Rewrite $121$ as $11^{2}$ .

$b = \frac{- 1 \pm \sqrt{11^{2}}}{2 \cdot 1}$

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

$b = \frac{- 1 \pm 11}{2 \cdot 1}$

Multiply $2$ by $1$ .

$b = \frac{- 1 \pm 11}{2}$

Step 4

The final answer is the combination of both solutions.

$b = 5, - 6$