Solve by Completing the Square 2x^2+16x+42=0

$2 x^{2} + 16 x + 42 = 0$

Step 1

Subtract $42$ from both sides of the equation.

$2 x^{2} + 16 x = - 42$

Step 2

Divide each term in $2 x^{2} + 16 x = - 42$ by $2$ and simplify.

Divide each term in $2 x^{2} + 16 x = - 42$ by $2$ .

$\frac{2 x^{2}}{2} + \frac{16 x}{2} = \frac{- 42}{2}$

Simplify the left side.

Simplify each term.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x^{2}}{2} + \frac{16 x}{2} = \frac{- 42}{2}$

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

$x^{2} + \frac{16 x}{2} = \frac{- 42}{2}$

Cancel the common factor of $16$ and $2$ .

Factor $2$ out of $16 x$ .

$x^{2} + \frac{2 (8 x)}{2} = \frac{- 42}{2}$

Cancel the common factors.

Factor $2$ out of $2$ .

$x^{2} + \frac{2 (8 x)}{2 (1)} = \frac{- 42}{2}$

Cancel the common factor.

$x^{2} + \frac{2 (8 x)}{2 \cdot 1} = \frac{- 42}{2}$

Rewrite the expression.

$x^{2} + \frac{8 x}{1} = \frac{- 42}{2}$

Divide $8 x$ by $1$ .

$x^{2} + 8 x = \frac{- 42}{2}$

Simplify the right side.

Divide $- 42$ by $2$ .

$x^{2} + 8 x = - 21$

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 $b$ .

${(\frac{b}{2})}^{2} = {(4)}^{2}$

Step 4

Add the term to each side of the equation.

$x^{2} + 8 x + {(4)}^{2} = - 21 + {(4)}^{2}$

Step 5

Simplify the equation.

Simplify the left side.

Raise $4$ to the power of $2$ .

$x^{2} + 8 x + 16 = - 21 + {(4)}^{2}$

Simplify the right side.

Simplify $- 21 + {(4)}^{2}$ .

Raise $4$ to the power of $2$ .

$x^{2} + 8 x + 16 = - 21 + 16$

Add $- 21$ and $16$ .

$x^{2} + 8 x + 16 = - 5$

Step 6

Factor the perfect trinomial square into ${(x + 4)}^{2}$ .

${(x + 4)}^{2} = - 5$

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.

$x + 4 = \pm \sqrt{- 5}$

Simplify $\pm \sqrt{- 5}$ .

Rewrite $- 5$ as $- 1 (5)$ .

$x + 4 = \pm \sqrt{- 1 (5)}$

Rewrite $\sqrt{- 1 (5)}$ as $\sqrt{- 1} \cdot \sqrt{5}$ .

$x + 4 = \pm \sqrt{- 1} \cdot \sqrt{5}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x + 4 = \pm i \sqrt{5}$

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

Subtract $4$ from both sides of the equation.

$x = \pm i \sqrt{5} - 4$

Reorder $\pm i \sqrt{5}$ and $- 4$ .

$x = - 4 \pm i \sqrt{5}$

Solve by Completing the Square 2x^2+16x+42=0

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