Solve Using the Quadratic Formula x^2-10x+34=0

$x^{2} - 10 x + 34 = 0$

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 $c = 34$ into the quadratic formula and solve for $x$ .

$\frac{10 \pm \sqrt{{(- 10)}^{2} - 4 \cdot (1 \cdot 34)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{10 \pm \sqrt{100 - 4 \cdot 1 \cdot 34}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 34$ .

Multiply $- 4$ by $1$ .

$x = \frac{10 \pm \sqrt{100 - 4 \cdot 34}}{2 \cdot 1}$

Multiply $- 4$ by $34$ .

$x = \frac{10 \pm \sqrt{100 - 136}}{2 \cdot 1}$

Subtract $136$ from $100$ .

$x = \frac{10 \pm \sqrt{- 36}}{2 \cdot 1}$

Rewrite $- 36$ as $- 1 (36)$ .

$x = \frac{10 \pm \sqrt{- 1 \cdot 36}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (36)}$ as $\sqrt{- 1} \cdot \sqrt{36}$ .

$x = \frac{10 \pm \sqrt{- 1} \cdot \sqrt{36}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{10 \pm i \cdot \sqrt{36}}{2 \cdot 1}$

Rewrite $36$ as $6^{2}$ .

$x = \frac{10 \pm i \cdot \sqrt{6^{2}}}{2 \cdot 1}$

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

$x = \frac{10 \pm i \cdot 6}{2 \cdot 1}$

Move $6$ to the left of $i$ .

$x = \frac{10 \pm 6 i}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{10 \pm 6 i}{2}$

Simplify $\frac{10 \pm 6 i}{2}$ .

$x = 5 \pm 3 i$