Solve for d |2d^2-3d|=5

$| 2 d^{2} - 3 d | = 5$

Step 1

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$2 d^{2} - 3 d = \pm 5$

Step 2

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.

$2 d^{2} - 3 d = 5$

Subtract $5$ from both sides of the equation.

$2 d^{2} - 3 d - 5 = 0$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 5 = - 10$ and whose sum is $b = - 3$ .

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

$2 d^{2} - 3 d - 5 = 0$

Rewrite $- 3$ as $2$ plus $- 5$

$2 d^{2} + (2 - 5) d - 5 = 0$

Apply the distributive property.

$2 d^{2} + 2 d - 5 d - 5 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 d^{2} + 2 d) - 5 d - 5 = 0$

Factor out the greatest common factor (GCF) from each group.

$2 d (d + 1) - 5 (d + 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $d + 1$ .

$(d + 1) (2 d - 5) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$d + 1 = 0$

$2 d - 5 = 0$

Set $d + 1$ equal to $0$ and solve for $d$ .

Set $d + 1$ equal to $0$ .

$d + 1 = 0$

Subtract $1$ from both sides of the equation.

$d = - 1$

Set $2 d - 5$ equal to $0$ and solve for $d$ .

Set $2 d - 5$ equal to $0$ .

$2 d - 5 = 0$

Solve $2 d - 5 = 0$ for $d$ .

Add $5$ to both sides of the equation.

$2 d = 5$

Divide each term in $2 d = 5$ by $2$ and simplify.

Divide each term in $2 d = 5$ by $2$ .

$\frac{2 d}{2} = \frac{5}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 d}{2} = \frac{5}{2}$

Divide $d$ by $1$ .

$d = \frac{5}{2}$

The final solution is all the values that make $(d + 1) (2 d - 5) = 0$ true.

$d = - 1, \frac{5}{2}$

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

$2 d^{2} - 3 d = - 5$

Add $5$ to both sides of the equation.

$2 d^{2} - 3 d + 5 = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 2$ , $b = - 3$ , and $c = 5$ into the quadratic formula and solve for $d$ .

$\frac{3 \pm \sqrt{{(- 3)}^{2} - 4 \cdot (2 \cdot 5)}}{2 \cdot 2}$

Simplify.

Simplify the numerator.

Raise $- 3$ to the power of $2$ .

$d = \frac{3 \pm \sqrt{9 - 4 \cdot 2 \cdot 5}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot 5$ .

Multiply $- 4$ by $2$ .

$d = \frac{3 \pm \sqrt{9 - 8 \cdot 5}}{2 \cdot 2}$

Multiply $- 8$ by $5$ .

$d = \frac{3 \pm \sqrt{9 - 40}}{2 \cdot 2}$

Subtract $40$ from $9$ .

$d = \frac{3 \pm \sqrt{- 31}}{2 \cdot 2}$

Rewrite $- 31$ as $- 1 (31)$ .

$d = \frac{3 \pm \sqrt{- 1 \cdot 31}}{2 \cdot 2}$

Rewrite $\sqrt{- 1 (31)}$ as $\sqrt{- 1} \cdot \sqrt{31}$ .

$d = \frac{3 \pm \sqrt{- 1} \cdot \sqrt{31}}{2 \cdot 2}$

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

$d = \frac{3 \pm i \sqrt{31}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$d = \frac{3 \pm i \sqrt{31}}{4}$

The final answer is the combination of both solutions.

$d = \frac{3 + i \sqrt{31}}{4}, \frac{3 - i \sqrt{31}}{4}$

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

$d = - 1, \frac{5}{2}, \frac{3 + i \sqrt{31}}{4}, \frac{3 - i \sqrt{31}}{4}$

Solve for d |2d^2-3d|=5

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