Solve Using the Quadratic Formula 0=x^2-x-6

$0 = x^{2} - x - 6$

Step 1

Move all the expressions to the left side of the equation.

Subtract $x^{2}$ from both sides of the equation.

$- x^{2} = - x - 6$

Add $x$ to both sides of the equation.

$- x^{2} + x = - 6$

Add $6$ to both sides of the equation.

$- x^{2} + x + 6 = 0$

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 $a = - 1$ , $b = 1$ , and $c = 6$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (- 1 \cdot 6)}}{2 \cdot - 1}$

Step 4

Simplify.

Simplify the numerator.

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot - 1 \cdot 6}}{2 \cdot - 1}$

Multiply $- 4 \cdot - 1 \cdot 6$ .

Multiply $- 4$ by $- 1$ .

$x = \frac{- 1 \pm \sqrt{1 + 4 \cdot 6}}{2 \cdot - 1}$

Multiply $4$ by $6$ .

$x = \frac{- 1 \pm \sqrt{1 + 24}}{2 \cdot - 1}$

Add $1$ and $24$ .

$x = \frac{- 1 \pm \sqrt{25}}{2 \cdot - 1}$

Rewrite $25$ as $5^{2}$ .

$x = \frac{- 1 \pm \sqrt{5^{2}}}{2 \cdot - 1}$

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

$x = \frac{- 1 \pm 5}{2 \cdot - 1}$

Multiply $2$ by $- 1$ .

$x = \frac{- 1 \pm 5}{- 2}$

Simplify $\frac{- 1 \pm 5}{- 2}$ .

$x = \frac{1 \pm 5}{2}$

Step 5

The final answer is the combination of both solutions.

$x = 3, - 2$