Solve Using the Quadratic Formula a^2+10a+24=0

$a^{2} + 10 a + 24 = 0$

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 $a = 1$ , $b = 10$ , and $c = 24$ into the quadratic formula and solve for $a$ .

$\frac{- 10 \pm \sqrt{10^{2} - 4 \cdot (1 \cdot 24)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $10$ to the power of $2$ .

$a = \frac{- 10 \pm \sqrt{100 - 4 \cdot 1 \cdot 24}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 24$ .

Multiply $- 4$ by $1$ .

$a = \frac{- 10 \pm \sqrt{100 - 4 \cdot 24}}{2 \cdot 1}$

Multiply $- 4$ by $24$ .

$a = \frac{- 10 \pm \sqrt{100 - 96}}{2 \cdot 1}$

Subtract $96$ from $100$ .

$a = \frac{- 10 \pm \sqrt{4}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

$a = \frac{- 10 \pm \sqrt{2^{2}}}{2 \cdot 1}$

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

$a = \frac{- 10 \pm 2}{2 \cdot 1}$

Multiply $2$ by $1$ .

$a = \frac{- 10 \pm 2}{2}$

Simplify $\frac{- 10 \pm 2}{2}$ .

$a = - 5 \pm 1$

Step 4

The final answer is the combination of both solutions.

$a = - 4, - 6$