Solve by Completing the Square 2y^2+10y=-7

$2 y^{2} + 10 y = - 7$

Step 1

Divide each term in $2 y^{2} + 10 y = - 7$ by $2$ and simplify.

Divide each term in $2 y^{2} + 10 y = - 7$ by $2$ .

$\frac{2 y^{2}}{2} + \frac{10 y}{2} = \frac{- 7}{2}$

Simplify the left side.

Simplify each term.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 y^{2}}{2} + \frac{10 y}{2} = \frac{- 7}{2}$

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

$y^{2} + \frac{10 y}{2} = \frac{- 7}{2}$

Cancel the common factor of $10$ and $2$ .

Factor $2$ out of $10 y$ .

$y^{2} + \frac{2 (5 y)}{2} = \frac{- 7}{2}$

Cancel the common factors.

Factor $2$ out of $2$ .

$y^{2} + \frac{2 (5 y)}{2 (1)} = \frac{- 7}{2}$

Cancel the common factor.

$y^{2} + \frac{2 (5 y)}{2 \cdot 1} = \frac{- 7}{2}$

Rewrite the expression.

$y^{2} + \frac{5 y}{1} = \frac{- 7}{2}$

Divide $5 y$ by $1$ .

$y^{2} + 5 y = \frac{- 7}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$y^{2} + 5 y = - \frac{7}{2}$

Step 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 $b$ .

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

Step 3

Add the term to each side of the equation.

$y^{2} + 5 y + {(\frac{5}{2})}^{2} = - \frac{7}{2} + {(\frac{5}{2})}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

Simplify each term.

Apply the product rule to $\frac{5}{2}$ .

$y^{2} + 5 y + \frac{5^{2}}{2^{2}} = - \frac{7}{2} + {(\frac{5}{2})}^{2}$

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

$y^{2} + 5 y + \frac{25}{2^{2}} = - \frac{7}{2} + {(\frac{5}{2})}^{2}$

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

$y^{2} + 5 y + \frac{25}{4} = - \frac{7}{2} + {(\frac{5}{2})}^{2}$

Simplify the right side.

Simplify $- \frac{7}{2} + {(\frac{5}{2})}^{2}$ .

Simplify each term.

Apply the product rule to $\frac{5}{2}$ .

$y^{2} + 5 y + \frac{25}{4} = - \frac{7}{2} + \frac{5^{2}}{2^{2}}$

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

$y^{2} + 5 y + \frac{25}{4} = - \frac{7}{2} + \frac{25}{2^{2}}$

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

$y^{2} + 5 y + \frac{25}{4} = - \frac{7}{2} + \frac{25}{4}$

To write $- \frac{7}{2}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$y^{2} + 5 y + \frac{25}{4} = - \frac{7}{2} \cdot \frac{2}{2} + \frac{25}{4}$

Write each expression with a common denominator of $4$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{7}{2}$ by $\frac{2}{2}$ .

$y^{2} + 5 y + \frac{25}{4} = - \frac{7 \cdot 2}{2 \cdot 2} + \frac{25}{4}$

Multiply $2$ by $2$ .

$y^{2} + 5 y + \frac{25}{4} = - \frac{7 \cdot 2}{4} + \frac{25}{4}$

Combine the numerators over the common denominator.

$y^{2} + 5 y + \frac{25}{4} = \frac{- 7 \cdot 2 + 25}{4}$

Simplify the numerator.

Multiply $- 7$ by $2$ .

$y^{2} + 5 y + \frac{25}{4} = \frac{- 14 + 25}{4}$

Add $- 14$ and $25$ .

$y^{2} + 5 y + \frac{25}{4} = \frac{11}{4}$

Step 5

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

${(y + \frac{5}{2})}^{2} = \frac{11}{4}$

Step 6

Solve the equation for $y$ .

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

$y + \frac{5}{2} = \pm \sqrt{\frac{11}{4}}$

Simplify $\pm \sqrt{\frac{11}{4}}$ .

Rewrite $\sqrt{\frac{11}{4}}$ as $\frac{\sqrt{11}}{\sqrt{4}}$ .

$y + \frac{5}{2} = \pm \frac{\sqrt{11}}{\sqrt{4}}$

Simplify the denominator.

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

$y + \frac{5}{2} = \pm \frac{\sqrt{11}}{\sqrt{2^{2}}}$

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

$y + \frac{5}{2} = \pm \frac{\sqrt{11}}{2}$

Subtract $\frac{5}{2}$ from both sides of the equation.

$y = \pm \frac{\sqrt{11}}{2} - \frac{5}{2}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$y = \pm \frac{\sqrt{11}}{2} - \frac{5}{2}$

Decimal Form:

$y = - 0.84168760 \dots, - 4.15831239 \dots$

Solve by Completing the Square 2y^2+10y=-7

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