Solve by Completing the Square 2x^2-6x+3=0

$2 x^{2} - 6 x + 3 = 0$

Step 1

Subtract $3$ from both sides of the equation.

$2 x^{2} - 6 x = - 3$

Step 2

Divide each term in $2 x^{2} - 6 x = - 3$ by $2$ and simplify.

Divide each term in $2 x^{2} - 6 x = - 3$ by $2$ .

$\frac{2 x^{2}}{2} + \frac{- 6 x}{2} = \frac{- 3}{2}$

Simplify the left side.

Simplify each term.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x^{2}}{2} + \frac{- 6 x}{2} = \frac{- 3}{2}$

Divide $x^{2}$ by $1$ .

$x^{2} + \frac{- 6 x}{2} = \frac{- 3}{2}$

Cancel the common factor of $- 6$ and $2$ .

Factor $2$ out of $- 6 x$ .

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

Cancel the common factors.

Factor $2$ out of $2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $- 3 x$ by $1$ .

$x^{2} - 3 x = \frac{- 3}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x^{2} - 3 x = - \frac{3}{2}$

Step 3

To create a trinomial square on the left side of the equation, find a value that is equal to the square of half of $b$ .

${(\frac{b}{2})}^{2} = {(- \frac{3}{2})}^{2}$

Step 4

Add the term to each side of the equation.

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

Step 5

Simplify the equation.

Simplify the left side.

Simplify each term.

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- \frac{3}{2}$ .

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

Apply the product rule to $\frac{3}{2}$ .

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

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

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

Multiply $\frac{3^{2}}{2^{2}}$ by $1$ .

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

Raise $3$ to the power of $2$ .

$x^{2} - 3 x + \frac{9}{2^{2}} = - \frac{3}{2} + {(- \frac{3}{2})}^{2}$

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

$x^{2} - 3 x + \frac{9}{4} = - \frac{3}{2} + {(- \frac{3}{2})}^{2}$

Simplify the right side.

Simplify $- \frac{3}{2} + {(- \frac{3}{2})}^{2}$ .

Simplify each term.

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- \frac{3}{2}$ .

$x^{2} - 3 x + \frac{9}{4} = - \frac{3}{2} + {(- 1)}^{2} {(\frac{3}{2})}^{2}$

Apply the product rule to $\frac{3}{2}$ .

$x^{2} - 3 x + \frac{9}{4} = - \frac{3}{2} + {(- 1)}^{2} \frac{3^{2}}{2^{2}}$

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

$x^{2} - 3 x + \frac{9}{4} = - \frac{3}{2} + 1 \frac{3^{2}}{2^{2}}$

Multiply $\frac{3^{2}}{2^{2}}$ by $1$ .

$x^{2} - 3 x + \frac{9}{4} = - \frac{3}{2} + \frac{3^{2}}{2^{2}}$

Raise $3$ to the power of $2$ .

$x^{2} - 3 x + \frac{9}{4} = - \frac{3}{2} + \frac{9}{2^{2}}$

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

$x^{2} - 3 x + \frac{9}{4} = - \frac{3}{2} + \frac{9}{4}$

To write $- \frac{3}{2}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$x^{2} - 3 x + \frac{9}{4} = - \frac{3}{2} \cdot \frac{2}{2} + \frac{9}{4}$

Write each expression with a common denominator of $4$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{3}{2}$ by $\frac{2}{2}$ .

$x^{2} - 3 x + \frac{9}{4} = - \frac{3 \cdot 2}{2 \cdot 2} + \frac{9}{4}$

Multiply $2$ by $2$ .

$x^{2} - 3 x + \frac{9}{4} = - \frac{3 \cdot 2}{4} + \frac{9}{4}$

Combine the numerators over the common denominator.

$x^{2} - 3 x + \frac{9}{4} = \frac{- 3 \cdot 2 + 9}{4}$

Simplify the numerator.

Multiply $- 3$ by $2$ .

$x^{2} - 3 x + \frac{9}{4} = \frac{- 6 + 9}{4}$

Add $- 6$ and $9$ .

$x^{2} - 3 x + \frac{9}{4} = \frac{3}{4}$

Step 6

Factor the perfect trinomial square into ${(x - \frac{3}{2})}^{2}$ .

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

Step 7

Solve the equation for $x$ .

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x - \frac{3}{2} = \pm \sqrt{\frac{3}{4}}$

Simplify $\pm \sqrt{\frac{3}{4}}$ .

Rewrite $\sqrt{\frac{3}{4}}$ as $\frac{\sqrt{3}}{\sqrt{4}}$ .

$x - \frac{3}{2} = \pm \frac{\sqrt{3}}{\sqrt{4}}$

Simplify the denominator.

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

$x - \frac{3}{2} = \pm \frac{\sqrt{3}}{\sqrt{2^{2}}}$

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

$x - \frac{3}{2} = \pm \frac{\sqrt{3}}{2}$

Add $\frac{3}{2}$ to both sides of the equation.

$x = \pm \frac{\sqrt{3}}{2} + \frac{3}{2}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$x = \pm \frac{\sqrt{3}}{2} + \frac{3}{2}$

Decimal Form:

$x = 2.36602540 \dots, 0.63397459 \dots$

Solve by Completing the Square 2x^2-6x+3=0

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