Solve Using the Quadratic Formula a^2+14a-51=0

$a^{2} + 14 a - 51 = 0$

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 $a = 1$ , $b = 14$ , and $c = - 51$ into the quadratic formula and solve for $a$ .

$\frac{- 14 \pm \sqrt{14^{2} - 4 \cdot (1 \cdot - 51)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $14$ to the power of $2$ .

$a = \frac{- 14 \pm \sqrt{196 - 4 \cdot 1 \cdot - 51}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 51$ .

Multiply $- 4$ by $1$ .

$a = \frac{- 14 \pm \sqrt{196 - 4 \cdot - 51}}{2 \cdot 1}$

Multiply $- 4$ by $- 51$ .

$a = \frac{- 14 \pm \sqrt{196 + 204}}{2 \cdot 1}$

Add $196$ and $204$ .

$a = \frac{- 14 \pm \sqrt{400}}{2 \cdot 1}$

Rewrite $400$ as $20^{2}$ .

$a = \frac{- 14 \pm \sqrt{20^{2}}}{2 \cdot 1}$

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

$a = \frac{- 14 \pm 20}{2 \cdot 1}$

Multiply $2$ by $1$ .

$a = \frac{- 14 \pm 20}{2}$

Simplify $\frac{- 14 \pm 20}{2}$ .

$a = - 7 \pm 10$

Step 4

The final answer is the combination of both solutions.

$a = 3, - 17$