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

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

Add $3$ to both sides of the equation.

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

Divide each term by $2$ and simplify.

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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 each term.

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Cancel the common factor of $2$ .

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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 $2$ .

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Factor $2$ out of $6 x$ .

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

Cancel the common factors.

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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 $1$ .

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

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

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

Add the term to each side of the equation.

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

Simplify the equation.

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Simplify each term.

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

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

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

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Simplify each term.

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

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

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 $1$ .

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

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

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.

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Multiply $3$ by $2$ .

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

Add $6$ and $9$ .

$x^{2} + 3 x + \frac{9}{4} = \frac{15}{4}$

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

${(x + \frac{3}{2})}^{2} = \frac{15}{4}$

Solve the equation for $x$ .

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Take the square root of each side of the equation to set up the solution for $x$

${(x + \frac{3}{2})}^{2 \cdot \frac{1}{2}} = \pm \sqrt{\frac{15}{4}}$

Remove the perfect root factor $x + \frac{3}{2}$ under the radical to solve for $x$ .

$x + \frac{3}{2} = \pm \sqrt{\frac{15}{4}}$

Simplify the right side of the equation.

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Rewrite $\sqrt{\frac{15}{4}}$ as $\frac{\sqrt{15}}{\sqrt{4}}$ .

$x + \frac{3}{2} = \pm \frac{\sqrt{15}}{\sqrt{4}}$

Simplify the denominator.

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

$x + \frac{3}{2} = \pm \frac{\sqrt{15}}{\sqrt{2^{2}}}$

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

$x + \frac{3}{2} = \pm \frac{\sqrt{15}}{2}$

Remove parentheses.

$x + \frac{3}{2} = \pm \frac{\sqrt{15}}{2}$

Subtract $\frac{3}{2}$ from both sides of the equation.

$x = \pm \frac{\sqrt{15}}{2} - \frac{3}{2}$

The result can be shown in multiple forms.

Exact Form:

$x = \pm \frac{\sqrt{15}}{2} - \frac{3}{2}$

Decimal Form:

$x = 0.43649167 \dots, - 3.43649167 \dots$

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

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