Solve by Completing the Square -3x^2-42x-51=-12x

$- 3 x^{2} - 42 x - 51 = - 12 x$

Step 1

Simplify the equation into a proper form to complete the square.

Add $51$ to both sides of the equation.

$- 3 x^{2} - 42 x = - 12 x + 51$

Move all terms containing $x$ to the left side of the equation.

Add $12 x$ to both sides of the equation.

$- 3 x^{2} - 42 x + 12 x = 51$

Add $- 42 x$ and $12 x$ .

$- 3 x^{2} - 30 x = 51$

Step 2

Divide each term in $- 3 x^{2} - 30 x = 51$ by $- 3$ and simplify.

Divide each term in $- 3 x^{2} - 30 x = 51$ by $- 3$ .

$\frac{- 3 x^{2}}{- 3} + \frac{- 30 x}{- 3} = \frac{51}{- 3}$

Simplify the left side.

Simplify each term.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 x^{2}}{- 3} + \frac{- 30 x}{- 3} = \frac{51}{- 3}$

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

$x^{2} + \frac{- 30 x}{- 3} = \frac{51}{- 3}$

Cancel the common factor of $- 30$ and $- 3$ .

Factor $- 3$ out of $- 30 x$ .

$x^{2} + \frac{- 3 (10 x)}{- 3} = \frac{51}{- 3}$

Cancel the common factors.

Factor $- 3$ out of $- 3$ .

$x^{2} + \frac{- 3 (10 x)}{- 3 (1)} = \frac{51}{- 3}$

Cancel the common factor.

$x^{2} + \frac{- 3 (10 x)}{- 3 \cdot 1} = \frac{51}{- 3}$

Rewrite the expression.

$x^{2} + \frac{10 x}{1} = \frac{51}{- 3}$

Divide $10 x$ by $1$ .

$x^{2} + 10 x = \frac{51}{- 3}$

Simplify the right side.

Divide $51$ by $- 3$ .

$x^{2} + 10 x = - 17$

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 $b$ .

${(\frac{b}{2})}^{2} = {(5)}^{2}$

Step 4

Add the term to each side of the equation.

$x^{2} + 10 x + {(5)}^{2} = - 17 + {(5)}^{2}$

Step 5

Simplify the equation.

Simplify the left side.

Raise $5$ to the power of $2$ .

$x^{2} + 10 x + 25 = - 17 + {(5)}^{2}$

Simplify the right side.

Simplify $- 17 + {(5)}^{2}$ .

Raise $5$ to the power of $2$ .

$x^{2} + 10 x + 25 = - 17 + 25$

Add $- 17$ and $25$ .

$x^{2} + 10 x + 25 = 8$

Step 6

Factor the perfect trinomial square into ${(x + 5)}^{2}$ .

${(x + 5)}^{2} = 8$

Step 7

Solve the equation for $x$ .

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

$x + 5 = \pm \sqrt{8}$

Simplify $\pm \sqrt{8}$ .

Rewrite $8$ as $2^{2} \cdot 2$ .

Factor $4$ out of $8$ .

$x + 5 = \pm \sqrt{4 (2)}$

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

$x + 5 = \pm \sqrt{2^{2} \cdot 2}$

Pull terms out from under the radical.

$x + 5 = \pm 2 \sqrt{2}$

Subtract $5$ from both sides of the equation.

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

Step 8

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = - 2.17157287 \dots, - 7.82842712 \dots$

Solve by Completing the Square -3x^2-42x-51=-12x

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