Solve for x 2x^2+x-1/2=0

$2 x^{2} + x - \frac{1}{2} = 0$

Multiply through by the least common denominator $2$ , then simplify.

Apply the distributive property.

$2 (2 x^{2}) + 2 x + 2 (- \frac{1}{2}) = 0$

Simplify.

Multiply $2$ by $2$ .

$4 x^{2} + 2 x + 2 (- \frac{1}{2}) = 0$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$4 x^{2} + 2 x + 2 (\frac{- 1}{2}) = 0$

Cancel the common factor.

$4 x^{2} + 2 x + 2 (\frac{- 1}{2}) = 0$

Rewrite the expression.

$4 x^{2} + 2 x - 1 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 4$ , $b = 2$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

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

Simplify.

Simplify the numerator.

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

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

Multiply $- 4$ by $4$ .

$x = \frac{- 2 \pm \sqrt{4 - 16 \cdot - 1}}{2 \cdot 4}$

Multiply $- 16$ by $- 1$ .

$x = \frac{- 2 \pm \sqrt{4 + 16}}{2 \cdot 4}$

Add $4$ and $16$ .

$x = \frac{- 2 \pm \sqrt{20}}{2 \cdot 4}$

Rewrite $20$ as $2^{2} \cdot 5$ .

Factor $4$ out of $20$ .

$x = \frac{- 2 \pm \sqrt{4 (5)}}{2 \cdot 4}$

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

$x = \frac{- 2 \pm \sqrt{2^{2} \cdot 5}}{2 \cdot 4}$

Pull terms out from under the radical.

$x = \frac{- 2 \pm 2 \sqrt{5}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{- 2 \pm 2 \sqrt{5}}{8}$

Simplify $\frac{- 2 \pm 2 \sqrt{5}}{8}$ .

$x = \frac{- 1 \pm \sqrt{5}}{4}$

The final answer is the combination of both solutions.

$x = - \frac{1 - \sqrt{5}}{4}, - \frac{1 + \sqrt{5}}{4}$

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{1 - \sqrt{5}}{4}, - \frac{1 + \sqrt{5}}{4}$

Decimal Form:

$x = 0.30901699 \dots, - 0.80901699 \dots$

Solve for x 2x^2+x-1/2=0

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