Solve Using the Quadratic Formula a^2-2a-35=0

$a^{2} - 2 a - 35 = 0$

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 $a = 1$ , $b = - 2$ , and $c = - 35$ into the quadratic formula and solve for $a$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (1 \cdot - 35)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

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

$a = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 35}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 35$ .

Multiply $- 4$ by $1$ .

$a = \frac{2 \pm \sqrt{4 - 4 \cdot - 35}}{2 \cdot 1}$

Multiply $- 4$ by $- 35$ .

$a = \frac{2 \pm \sqrt{4 + 140}}{2 \cdot 1}$

Add $4$ and $140$ .

$a = \frac{2 \pm \sqrt{144}}{2 \cdot 1}$

Rewrite $144$ as $12^{2}$ .

$a = \frac{2 \pm \sqrt{12^{2}}}{2 \cdot 1}$

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

$a = \frac{2 \pm 12}{2 \cdot 1}$

Multiply $2$ by $1$ .

$a = \frac{2 \pm 12}{2}$

Simplify $\frac{2 \pm 12}{2}$ .

$a = 1 \pm 6$

Step 4

The final answer is the combination of both solutions.

$a = 7, - 5$