Solve by Completing the Square x^2-3x-12=0

$x^{2} - 3 x - 12 = 0$

Add $12$ to both sides of the equation.

$x^{2} - 3 x = 12$

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.

$x^{2} - 3 x + {(- \frac{3}{2})}^{2} = 12 + {(- \frac{3}{2})}^{2}$

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 $- \frac{3}{2}$ .

$x^{2} - 3 x + {(- 1)}^{2} {(\frac{3}{2})}^{2} = 12 + {(- \frac{3}{2})}^{2}$

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

$x^{2} - 3 x + {(- 1)}^{2} \frac{3^{2}}{2^{2}} = 12 + {(- \frac{3}{2})}^{2}$

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

$x^{2} - 3 x + 1 \frac{3^{2}}{2^{2}} = 12 + {(- \frac{3}{2})}^{2}$

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

$x^{2} - 3 x + \frac{3^{2}}{2^{2}} = 12 + {(- \frac{3}{2})}^{2}$

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

$x^{2} - 3 x + \frac{9}{2^{2}} = 12 + {(- \frac{3}{2})}^{2}$

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

$x^{2} - 3 x + \frac{9}{4} = 12 + {(- \frac{3}{2})}^{2}$

Simplify the right side.

Simplify $12 + {(- \frac{3}{2})}^{2}$ .

Simplify each term.

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

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

$x^{2} - 3 x + \frac{9}{4} = 12 + {(- 1)}^{2} {(\frac{3}{2})}^{2}$

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

$x^{2} - 3 x + \frac{9}{4} = 12 + {(- 1)}^{2} \frac{3^{2}}{2^{2}}$

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

$x^{2} - 3 x + \frac{9}{4} = 12 + 1 \frac{3^{2}}{2^{2}}$

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

$x^{2} - 3 x + \frac{9}{4} = 12 + \frac{3^{2}}{2^{2}}$

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

$x^{2} - 3 x + \frac{9}{4} = 12 + \frac{9}{2^{2}}$

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

$x^{2} - 3 x + \frac{9}{4} = 12 + \frac{9}{4}$

To write $12$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$x^{2} - 3 x + \frac{9}{4} = 12 \cdot \frac{4}{4} + \frac{9}{4}$

Combine $12$ and $\frac{4}{4}$ .

$x^{2} - 3 x + \frac{9}{4} = \frac{12 \cdot 4}{4} + \frac{9}{4}$

Combine the numerators over the common denominator.

$x^{2} - 3 x + \frac{9}{4} = \frac{12 \cdot 4 + 9}{4}$

Simplify the numerator.

Multiply $12$ by $4$ .

$x^{2} - 3 x + \frac{9}{4} = \frac{48 + 9}{4}$

Add $48$ and $9$ .

$x^{2} - 3 x + \frac{9}{4} = \frac{57}{4}$

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

${(x - \frac{3}{2})}^{2} = \frac{57}{4}$

Solve the equation for $x$ .

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

$x - \frac{3}{2} = \pm \sqrt{\frac{57}{4}}$

Simplify $\pm \sqrt{\frac{57}{4}}$ .

Rewrite $\sqrt{\frac{57}{4}}$ as $\frac{\sqrt{57}}{\sqrt{4}}$ .

$x - \frac{3}{2} = \pm \frac{\sqrt{57}}{\sqrt{4}}$

Simplify the denominator.

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

$x - \frac{3}{2} = \pm \frac{\sqrt{57}}{\sqrt{2^{2}}}$

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

$x - \frac{3}{2} = \pm \frac{\sqrt{57}}{2}$

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

$x = \pm \frac{\sqrt{57}}{2} + \frac{3}{2}$

The result can be shown in multiple forms.

Exact Form:

$x = \pm \frac{\sqrt{57}}{2} + \frac{3}{2}$

Decimal Form:

$x = 5.27491721 \dots, - 2.27491721 \dots$

Solve by Completing the Square x^2-3x-12=0

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