Solve for d 3d^2+8d=40

$3 d^{2} + 8 d = 40$

Step 1

Subtract $40$ from both sides of the equation.

$3 d^{2} + 8 d - 40 = 0$

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 $a = 3$ , $b = 8$ , and $c = - 40$ into the quadratic formula and solve for $d$ .

$\frac{- 8 \pm \sqrt{8^{2} - 4 \cdot (3 \cdot - 40)}}{2 \cdot 3}$

Step 4

Simplify.

Simplify the numerator.

Raise $8$ to the power of $2$ .

$d = \frac{- 8 \pm \sqrt{64 - 4 \cdot 3 \cdot - 40}}{2 \cdot 3}$

Multiply $- 4 \cdot 3 \cdot - 40$ .

Multiply $- 4$ by $3$ .

$d = \frac{- 8 \pm \sqrt{64 - 12 \cdot - 40}}{2 \cdot 3}$

Multiply $- 12$ by $- 40$ .

$d = \frac{- 8 \pm \sqrt{64 + 480}}{2 \cdot 3}$

Add $64$ and $480$ .

$d = \frac{- 8 \pm \sqrt{544}}{2 \cdot 3}$

Rewrite $544$ as $4^{2} \cdot 34$ .

Factor $16$ out of $544$ .

$d = \frac{- 8 \pm \sqrt{16 (34)}}{2 \cdot 3}$

Rewrite $16$ as $4^{2}$ .

$d = \frac{- 8 \pm \sqrt{4^{2} \cdot 34}}{2 \cdot 3}$

Pull terms out from under the radical.

$d = \frac{- 8 \pm 4 \sqrt{34}}{2 \cdot 3}$

Multiply $2$ by $3$ .

$d = \frac{- 8 \pm 4 \sqrt{34}}{6}$

Simplify $\frac{- 8 \pm 4 \sqrt{34}}{6}$ .

$d = \frac{- 4 \pm 2 \sqrt{34}}{3}$

Step 5

The final answer is the combination of both solutions.

$d = - \frac{4 - 2 \sqrt{34}}{3}, - \frac{4 + 2 \sqrt{34}}{3}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$d = - \frac{4 - 2 \sqrt{34}}{3}, - \frac{4 + 2 \sqrt{34}}{3}$

Decimal Form:

$d = 2.55396792 \dots, - 5.22063459 \dots$