Solve Using the Quadratic Formula 2d^2+4=5d

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

Step 1

Subtract $5 d$ from both sides of the equation.

$2 d^{2} + 4 - 5 d = 0$

Step 2

Use the quadratic formula to find the solutions.

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

Step 3

Substitute the values $a = 2$ , $b = - 5$ , and $c = 4$ into the quadratic formula and solve for $d$ .

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

Step 4

Simplify.

Simplify the numerator.

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

$d = \frac{5 \pm \sqrt{25 - 4 \cdot 2 \cdot 4}}{2 \cdot 2}$

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

Multiply $- 4$ by $2$ .

$d = \frac{5 \pm \sqrt{25 - 8 \cdot 4}}{2 \cdot 2}$

Multiply $- 8$ by $4$ .

$d = \frac{5 \pm \sqrt{25 - 32}}{2 \cdot 2}$

Subtract $32$ from $25$ .

$d = \frac{5 \pm \sqrt{- 7}}{2 \cdot 2}$

Rewrite $- 7$ as $- 1 (7)$ .

$d = \frac{5 \pm \sqrt{- 1 \cdot 7}}{2 \cdot 2}$

Rewrite $\sqrt{- 1 (7)}$ as $\sqrt{- 1} \cdot \sqrt{7}$ .

$d = \frac{5 \pm \sqrt{- 1} \cdot \sqrt{7}}{2 \cdot 2}$

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

$d = \frac{5 \pm i \sqrt{7}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$d = \frac{5 \pm i \sqrt{7}}{4}$