Solve for d 5d^2+19d+16=0

$5 d^{2} + 19 d + 16 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 5$ , $b = 19$ , and $c = 16$ into the quadratic formula and solve for $d$ .

$\frac{- 19 \pm \sqrt{19^{2} - 4 \cdot (5 \cdot 16)}}{2 \cdot 5}$

Simplify.

Simplify the numerator.

Raise $19$ to the power of $2$ .

$d = \frac{- 19 \pm \sqrt{361 - 4 \cdot 5 \cdot 16}}{2 \cdot 5}$

Multiply $- 4$ by $5$ .

$d = \frac{- 19 \pm \sqrt{361 - 20 \cdot 16}}{2 \cdot 5}$

Multiply $- 20$ by $16$ .

$d = \frac{- 19 \pm \sqrt{361 - 320}}{2 \cdot 5}$

Subtract $320$ from $361$ .

$d = \frac{- 19 \pm \sqrt{41}}{2 \cdot 5}$

Multiply $2$ by $5$ .

$d = \frac{- 19 \pm \sqrt{41}}{10}$

The final answer is the combination of both solutions.

$d = - \frac{19 - \sqrt{41}}{10}, - \frac{19 + \sqrt{41}}{10}$

The result can be shown in multiple forms.

Exact Form:

$d = - \frac{19 - \sqrt{41}}{10}, - \frac{19 + \sqrt{41}}{10}$

Decimal Form:

$d = - 1.25968757 \dots, - 2.54031242 \dots$