Solve Using the Quadratic Formula m^2+10m+14=-7

$m^{2} + 10 m + 14 = - 7$

Step 1

Move all terms to the left side of the equation and simplify.

Add $7$ to both sides of the equation.

$m^{2} + 10 m + 14 + 7 = 0$

Add $14$ and $7$ .

$m^{2} + 10 m + 21 = 0$

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 $a = 1$ , $b = 10$ , and $c = 21$ into the quadratic formula and solve for $m$ .

$\frac{- 10 \pm \sqrt{10^{2} - 4 \cdot (1 \cdot 21)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

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

$m = \frac{- 10 \pm \sqrt{100 - 4 \cdot 1 \cdot 21}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 21$ .

Multiply $- 4$ by $1$ .

$m = \frac{- 10 \pm \sqrt{100 - 4 \cdot 21}}{2 \cdot 1}$

Multiply $- 4$ by $21$ .

$m = \frac{- 10 \pm \sqrt{100 - 84}}{2 \cdot 1}$

Subtract $84$ from $100$ .

$m = \frac{- 10 \pm \sqrt{16}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

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

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

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

Multiply $2$ by $1$ .

$m = \frac{- 10 \pm 4}{2}$

Simplify $\frac{- 10 \pm 4}{2}$ .

$m = - 5 \pm 2$

Step 5

The final answer is the combination of both solutions.

$m = - 3, - 7$