Solve Using the Quadratic Formula a^2+11a+28=0

$a^{2} + 11 a + 28 = 0$

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 $a = 1$ , $b = 11$ , and $c = 28$ into the quadratic formula and solve for $a$ .

$\frac{- 11 \pm \sqrt{11^{2} - 4 \cdot (1 \cdot 28)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $11$ to the power of $2$ .

$a = \frac{- 11 \pm \sqrt{121 - 4 \cdot 1 \cdot 28}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 28$ .

Multiply $- 4$ by $1$ .

$a = \frac{- 11 \pm \sqrt{121 - 4 \cdot 28}}{2 \cdot 1}$

Multiply $- 4$ by $28$ .

$a = \frac{- 11 \pm \sqrt{121 - 112}}{2 \cdot 1}$

Subtract $112$ from $121$ .

$a = \frac{- 11 \pm \sqrt{9}}{2 \cdot 1}$

Rewrite $9$ as $3^{2}$ .

$a = \frac{- 11 \pm \sqrt{3^{2}}}{2 \cdot 1}$

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

$a = \frac{- 11 \pm 3}{2 \cdot 1}$

Multiply $2$ by $1$ .

$a = \frac{- 11 \pm 3}{2}$

Step 4

The final answer is the combination of both solutions.

$a = - 4, - 7$