Solve Using the Quadratic Formula 0=x^2-2x-15

$0 = x^{2} - 2 x - 15$

Step 1

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

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

$- x^{2} = - 2 x - 15$

Add $2 x$ to both sides of the equation.

$- x^{2} + 2 x = - 15$

Add $15$ to both sides of the equation.

$- x^{2} + 2 x + 15 = 0$

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 $a = - 1$ , $b = 2$ , and $c = 15$ into the quadratic formula and solve for $x$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (- 1 \cdot 15)}}{2 \cdot - 1}$

Step 4

Simplify.

Simplify the numerator.

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 1 \cdot 15}}{2 \cdot - 1}$

Multiply $- 4 \cdot - 1 \cdot 15$ .

Multiply $- 4$ by $- 1$ .

$x = \frac{- 2 \pm \sqrt{4 + 4 \cdot 15}}{2 \cdot - 1}$

Multiply $4$ by $15$ .

$x = \frac{- 2 \pm \sqrt{4 + 60}}{2 \cdot - 1}$

Add $4$ and $60$ .

$x = \frac{- 2 \pm \sqrt{64}}{2 \cdot - 1}$

Rewrite $64$ as $8^{2}$ .

$x = \frac{- 2 \pm \sqrt{8^{2}}}{2 \cdot - 1}$

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

$x = \frac{- 2 \pm 8}{2 \cdot - 1}$

Multiply $2$ by $- 1$ .

$x = \frac{- 2 \pm 8}{- 2}$

Simplify $\frac{- 2 \pm 8}{- 2}$ .

$x = 1 \pm 4$

Step 5

The final answer is the combination of both solutions.

$x = 5, - 3$