Find the Roots (Zeros) 2x^2+6=x

$2 x^{2} + 6 = x$

Step 1

Subtract $x$ from both sides of the equation.

$2 x^{2} + 6 - x = 0$

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 $a = 2$ , $b = - 1$ , and $c = 6$ into the quadratic formula and solve for $x$ .

$\frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \cdot (2 \cdot 6)}}{2 \cdot 2}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 1$ to the power of $2$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 2 \cdot 6}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot 6$ .

Multiply $- 4$ by $2$ .

$x = \frac{1 \pm \sqrt{1 - 8 \cdot 6}}{2 \cdot 2}$

Multiply $- 8$ by $6$ .

$x = \frac{1 \pm \sqrt{1 - 48}}{2 \cdot 2}$

Subtract $48$ from $1$ .

$x = \frac{1 \pm \sqrt{- 47}}{2 \cdot 2}$

Rewrite $- 47$ as $- 1 (47)$ .

$x = \frac{1 \pm \sqrt{- 1 \cdot 47}}{2 \cdot 2}$

Rewrite $\sqrt{- 1 (47)}$ as $\sqrt{- 1} \cdot \sqrt{47}$ .

$x = \frac{1 \pm \sqrt{- 1} \cdot \sqrt{47}}{2 \cdot 2}$

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

$x = \frac{1 \pm i \sqrt{47}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{1 \pm i \sqrt{47}}{4}$

Step 5