Solve by Completing the Square 3x^2-6x-45=0

$3 x^{2} - 6 x - 45 = 0$

Step 1

Add $45$ to both sides of the equation.

$3 x^{2} - 6 x = 45$

Step 2

Divide each term in $3 x^{2} - 6 x = 45$ by $3$ and simplify.

Divide each term in $3 x^{2} - 6 x = 45$ by $3$ .

$\frac{3 x^{2}}{3} + \frac{- 6 x}{3} = \frac{45}{3}$

Simplify the left side.

Simplify each term.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x^{2}}{3} + \frac{- 6 x}{3} = \frac{45}{3}$

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

$x^{2} + \frac{- 6 x}{3} = \frac{45}{3}$

Cancel the common factor of $- 6$ and $3$ .

Factor $3$ out of $- 6 x$ .

$x^{2} + \frac{3 (- 2 x)}{3} = \frac{45}{3}$

Cancel the common factors.

Factor $3$ out of $3$ .

$x^{2} + \frac{3 (- 2 x)}{3 (1)} = \frac{45}{3}$

Cancel the common factor.

$x^{2} + \frac{3 (- 2 x)}{3 \cdot 1} = \frac{45}{3}$

Rewrite the expression.

$x^{2} + \frac{- 2 x}{1} = \frac{45}{3}$

Divide $- 2 x$ by $1$ .

$x^{2} - 2 x = \frac{45}{3}$

Simplify the right side.

Divide $45$ by $3$ .

$x^{2} - 2 x = 15$

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 $b$ .

${(\frac{b}{2})}^{2} = {(- 1)}^{2}$

Step 4

Add the term to each side of the equation.

$x^{2} - 2 x + {(- 1)}^{2} = 15 + {(- 1)}^{2}$

Step 5

Simplify the equation.

Simplify the left side.

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

$x^{2} - 2 x + 1 = 15 + {(- 1)}^{2}$

Simplify the right side.

Simplify $15 + {(- 1)}^{2}$ .

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

$x^{2} - 2 x + 1 = 15 + 1$

Add $15$ and $1$ .

$x^{2} - 2 x + 1 = 16$

Step 6

Factor the perfect trinomial square into ${(x - 1)}^{2}$ .

${(x - 1)}^{2} = 16$

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.

$x - 1 = \pm \sqrt{16}$

Simplify $\pm \sqrt{16}$ .

Rewrite $16$ as $4^{2}$ .

$x - 1 = \pm \sqrt{4^{2}}$

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

$x - 1 = \pm 4$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x - 1 = 4$

Move all terms not containing $x$ to the right side of the equation.

Add $1$ to both sides of the equation.

$x = 4 + 1$

Add $4$ and $1$ .

$x = 5$

Next, use the negative value of the $\pm$ to find the second solution.

$x - 1 = - 4$

Move all terms not containing $x$ to the right side of the equation.

Add $1$ to both sides of the equation.

$x = - 4 + 1$

Add $- 4$ and $1$ .

$x = - 3$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 5, - 3$

Solve by Completing the Square 3x^2-6x-45=0

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