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

$a^{2} - 2 a + 10 = 0$

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 $a = 1$ , $b = - 2$ , and $c = 10$ into the quadratic formula and solve for $a$ .

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

Step 3

Simplify.

Simplify the numerator.

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

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

Multiply $- 4 \cdot 1 \cdot 10$ .

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $10$ .

$a = \frac{2 \pm \sqrt{4 - 40}}{2 \cdot 1}$

Subtract $40$ from $4$ .

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

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

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

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

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

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

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

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

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

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

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

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

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

Multiply $2$ by $1$ .

$a = \frac{2 \pm 6 i}{2}$

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

$a = 1 \pm 3 i$