Solve Using the Quadratic Formula 2x^2-3x+1/2=0

$2 x^{2} - 3 x + \frac{1}{2} = 0$

Step 1

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

Apply the distributive property.

$2 (2 x^{2}) + 2 (- 3 x) + 2 (\frac{1}{2}) = 0$

Simplify.

Multiply $2$ by $2$ .

$4 x^{2} + 2 (- 3 x) + 2 (\frac{1}{2}) = 0$

Multiply $- 3$ by $2$ .

$4 x^{2} - 6 x + 2 (\frac{1}{2}) = 0$

Cancel the common factor of $2$ .

Cancel the common factor.

$4 x^{2} - 6 x + 2 (\frac{1}{2}) = 0$

Rewrite the expression.

$4 x^{2} - 6 x + 1 = 0$

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 $a = 4$ , $b = - 6$ , and $c = 1$ into the quadratic formula and solve for $x$ .

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

Step 4

Simplify.

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 4 \cdot 1}}{2 \cdot 4}$

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

Multiply $- 4$ by $4$ .

$x = \frac{6 \pm \sqrt{36 - 16 \cdot 1}}{2 \cdot 4}$

Multiply $- 16$ by $1$ .

$x = \frac{6 \pm \sqrt{36 - 16}}{2 \cdot 4}$

Subtract $16$ from $36$ .

$x = \frac{6 \pm \sqrt{20}}{2 \cdot 4}$

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

Factor $4$ out of $20$ .

$x = \frac{6 \pm \sqrt{4 (5)}}{2 \cdot 4}$

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

$x = \frac{6 \pm \sqrt{2^{2} \cdot 5}}{2 \cdot 4}$

Pull terms out from under the radical.

$x = \frac{6 \pm 2 \sqrt{5}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{6 \pm 2 \sqrt{5}}{8}$

Simplify $\frac{6 \pm 2 \sqrt{5}}{8}$ .

$x = \frac{3 \pm \sqrt{5}}{4}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \frac{3 \pm \sqrt{5}}{4}$

Decimal Form:

$x = 1.30901699 \dots, 0.19098300 \dots$

Solve Using the Quadratic Formula 2x^2-3x+1/2=0

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