Solve Using the Quadratic Formula x^2+x-182=0

$x^{2} + x - 182 = 0$

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 $c = - 182$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (1 \cdot - 182)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot - 182}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 182$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot - 182}}{2 \cdot 1}$

Multiply $- 4$ by $- 182$ .

$x = \frac{- 1 \pm \sqrt{1 + 728}}{2 \cdot 1}$

Add $1$ and $728$ .

$x = \frac{- 1 \pm \sqrt{729}}{2 \cdot 1}$

Rewrite $729$ as $27^{2}$ .

$x = \frac{- 1 \pm \sqrt{27^{2}}}{2 \cdot 1}$

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

$x = \frac{- 1 \pm 27}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm 27}{2}$

Step 4

The final answer is the combination of both solutions.

$x = 13, - 14$