Solve Using the Quadratic Formula 6+2x^2=20

$6 + 2 x^{2} = 20$

Move all terms to the left side of the equation and simplify.

Subtract $20$ from both sides of the equation.

$6 + 2 x^{2} - 20 = 0$

Subtract $20$ from $6$ .

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

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 2$ , $b = 0$ , and $c = - 14$ into the quadratic formula and solve for $x$ .

$\frac{0 \pm \sqrt{0^{2} - 4 \cdot (2 \cdot - 14)}}{2 \cdot 2}$

Simplify.

Simplify the numerator.

Raising $0$ to any positive power yields $0$ .

$x = \frac{0 \pm \sqrt{0 - 4 \cdot 2 \cdot - 14}}{2 \cdot 2}$

Multiply $- 4$ by $2$ .

$x = \frac{0 \pm \sqrt{0 - 8 \cdot - 14}}{2 \cdot 2}$

Multiply $- 8$ by $- 14$ .

$x = \frac{0 \pm \sqrt{0 + 112}}{2 \cdot 2}$

Add $0$ and $112$ .

$x = \frac{0 \pm \sqrt{112}}{2 \cdot 2}$

Rewrite $112$ as $4^{2} \cdot 7$ .

Factor $16$ out of $112$ .

$x = \frac{0 \pm \sqrt{16 (7)}}{2 \cdot 2}$

Rewrite $16$ as $4^{2}$ .

$x = \frac{0 \pm \sqrt{4^{2} \cdot 7}}{2 \cdot 2}$

Pull terms out from under the radical.

$x = \frac{0 \pm 4 \sqrt{7}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{0 \pm 4 \sqrt{7}}{4}$

Simplify $\frac{0 \pm 4 \sqrt{7}}{4}$ .

$x = \pm \sqrt{7}$

The result can be shown in multiple forms.

Exact Form:

$x = \pm \sqrt{7}$

Decimal Form:

$x = 2.64575131 \dots, - 2.64575131 \dots$