Solve Using the Quadratic Formula x^2-14x+49=0

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

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 $a = 1$ , $b = - 14$ , 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 3

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $49$ .

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

Subtract $196$ from $196$ .

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

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

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

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

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

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

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

Multiply $2$ by $1$ .

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

Divide $14$ by $2$ .

$x = 7$

Step 4

The final answer is the combination of both solutions.

$x = 7$ Double roots