Algebra Examples

$x^{2} + 9 x + 14 = 0$

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 $a = 1$ , $b = 9$ , and $c = 14$ into the quadratic formula and solve for $x$ .

$\frac{- 9 \pm \sqrt{9^{2} - 4 \cdot (1 \cdot 14)}}{2 \cdot 1}$

Step 3

Simplify.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Raise $9$ to the power of $2$ .

$x = \frac{- 9 \pm \sqrt{81 - 4 \cdot 1 \cdot 14}}{2 \cdot 1}$

Step 3.1.2

Multiply $- 4 \cdot 1 \cdot 14$ .

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Step 3.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 9 \pm \sqrt{81 - 4 \cdot 14}}{2 \cdot 1}$

Step 3.1.2.2

Multiply $- 4$ by $14$ .

$x = \frac{- 9 \pm \sqrt{81 - 56}}{2 \cdot 1}$

Step 3.1.3

Subtract $56$ from $81$ .

$x = \frac{- 9 \pm \sqrt{25}}{2 \cdot 1}$

Step 3.1.4

Rewrite $25$ as $5^{2}$ .

$x = \frac{- 9 \pm \sqrt{5^{2}}}{2 \cdot 1}$

Step 3.1.5

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

$x = \frac{- 9 \pm 5}{2 \cdot 1}$

Step 3.2

Multiply $2$ by $1$ .

$x = \frac{- 9 \pm 5}{2}$

Step 4

The final answer is the combination of both solutions.

$x = - 2, - 7$