Algebra Examples

$x^{2} - 14 x + 49$

Step 1

Set $x^{2} - 14 x + 49$ equal to $0$ .

$x^{2} - 14 x + 49 = 0$

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 $a = 1$ , $b = - 14$ , and $c = 49$ into the quadratic formula and solve for $x$ .

$\frac{14 \pm \sqrt{{(- 14)}^{2} - 4 \cdot (1 \cdot 49)}}{2 \cdot 1}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

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

$x = \frac{14 \pm \sqrt{196 - 4 \cdot 1 \cdot 49}}{2 \cdot 1}$

Step 4.1.2

Multiply $- 4 \cdot 1 \cdot 49$ .

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Step 4.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{14 \pm \sqrt{196 - 4 \cdot 49}}{2 \cdot 1}$

Step 4.1.2.2

Multiply $- 4$ by $49$ .

$x = \frac{14 \pm \sqrt{196 - 196}}{2 \cdot 1}$

Step 4.1.3

Subtract $196$ from $196$ .

$x = \frac{14 \pm \sqrt{0}}{2 \cdot 1}$

Step 4.1.4

Rewrite $0$ as $0^{2}$ .

$x = \frac{14 \pm \sqrt{0^{2}}}{2 \cdot 1}$

Step 4.1.5

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

$x = \frac{14 \pm 0}{2 \cdot 1}$

Step 4.1.6

$14$ plus or minus $0$ is $14$ .

$x = \frac{14}{2 \cdot 1}$

Step 4.2

Multiply $2$ by $1$ .

$x = \frac{14}{2}$

Step 4.3

Divide $14$ by $2$ .

$x = 7$

Step 5

The final answer is the combination of both solutions.

$x = 7$ Double roots