Solve by Completing the Square 2d^2=8+10d

$2 d^{2} = 8 + 10 d$

Step 1

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

Reorder $8$ and $10 d$ .

$2 d^{2} = 10 d + 8$

Subtract $10 d$ from both sides of the equation.

$2 d^{2} - 10 d = 8$

Step 2

Divide each term in $2 d^{2} - 10 d = 8$ by $2$ and simplify.

Divide each term in $2 d^{2} - 10 d = 8$ by $2$ .

$\frac{2 d^{2}}{2} + \frac{- 10 d}{2} = \frac{8}{2}$

Simplify the left side.

Simplify each term.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 d^{2}}{2} + \frac{- 10 d}{2} = \frac{8}{2}$

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

$d^{2} + \frac{- 10 d}{2} = \frac{8}{2}$

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

Factor $2$ out of $- 10 d$ .

$d^{2} + \frac{2 (- 5 d)}{2} = \frac{8}{2}$

Cancel the common factors.

Factor $2$ out of $2$ .

$d^{2} + \frac{2 (- 5 d)}{2 (1)} = \frac{8}{2}$

Cancel the common factor.

$d^{2} + \frac{2 (- 5 d)}{2 \cdot 1} = \frac{8}{2}$

Rewrite the expression.

$d^{2} + \frac{- 5 d}{1} = \frac{8}{2}$

Divide $- 5 d$ by $1$ .

$d^{2} - 5 d = \frac{8}{2}$

Simplify the right side.

Divide $8$ by $2$ .

$d^{2} - 5 d = 4$

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 $b$ .

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

Step 4

Add the term to each side of the equation.

$d^{2} - 5 d + {(- \frac{5}{2})}^{2} = 4 + {(- \frac{5}{2})}^{2}$

Step 5

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 $- \frac{5}{2}$ .

$d^{2} - 5 d + {(- 1)}^{2} {(\frac{5}{2})}^{2} = 4 + {(- \frac{5}{2})}^{2}$

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

$d^{2} - 5 d + {(- 1)}^{2} \frac{5^{2}}{2^{2}} = 4 + {(- \frac{5}{2})}^{2}$

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

$d^{2} - 5 d + 1 \frac{5^{2}}{2^{2}} = 4 + {(- \frac{5}{2})}^{2}$

Multiply $\frac{5^{2}}{2^{2}}$ by $1$ .

$d^{2} - 5 d + \frac{5^{2}}{2^{2}} = 4 + {(- \frac{5}{2})}^{2}$

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

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

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

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

Simplify the right side.

Simplify $4 + {(- \frac{5}{2})}^{2}$ .

Simplify each term.

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

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

$d^{2} - 5 d + \frac{25}{4} = 4 + {(- 1)}^{2} {(\frac{5}{2})}^{2}$

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

$d^{2} - 5 d + \frac{25}{4} = 4 + {(- 1)}^{2} \frac{5^{2}}{2^{2}}$

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

$d^{2} - 5 d + \frac{25}{4} = 4 + 1 \frac{5^{2}}{2^{2}}$

Multiply $\frac{5^{2}}{2^{2}}$ by $1$ .

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

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

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

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

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

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

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $4$ by $4$ .

$d^{2} - 5 d + \frac{25}{4} = \frac{16 + 25}{4}$

Add $16$ and $25$ .

$d^{2} - 5 d + \frac{25}{4} = \frac{41}{4}$

Step 6

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

${(d - \frac{5}{2})}^{2} = \frac{41}{4}$

Step 7

Solve the equation for $d$ .

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

$d - \frac{5}{2} = \pm \sqrt{\frac{41}{4}}$

Simplify $\pm \sqrt{\frac{41}{4}}$ .

Rewrite $\sqrt{\frac{41}{4}}$ as $\frac{\sqrt{41}}{\sqrt{4}}$ .

$d - \frac{5}{2} = \pm \frac{\sqrt{41}}{\sqrt{4}}$

Simplify the denominator.

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

$d - \frac{5}{2} = \pm \frac{\sqrt{41}}{\sqrt{2^{2}}}$

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

$d - \frac{5}{2} = \pm \frac{\sqrt{41}}{2}$

Add $\frac{5}{2}$ to both sides of the equation.

$d = \pm \frac{\sqrt{41}}{2} + \frac{5}{2}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$d = \pm \frac{\sqrt{41}}{2} + \frac{5}{2}$

Decimal Form:

$d = 5.70156211 \dots, - 0.70156211 \dots$

Solve by Completing the Square 2d^2=8+10d

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