Find All Complex Solutions 2x^2+16x-10=0

$2 x^{2} + 16 x - 10 = 0$

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

Substitute the values $a = 2$ , $b = 16$ , and $c = - 10$ into the quadratic formula and solve for $x$ .

$\frac{- 16 \pm \sqrt{16^{2} - 4 \cdot (2 \cdot - 10)}}{2 \cdot 2}$

Step 3

Simplify.

Simplify the numerator.

Raise $16$ to the power of $2$ .

$x = \frac{- 16 \pm \sqrt{256 - 4 \cdot 2 \cdot - 10}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 10$ .

Multiply $- 4$ by $2$ .

$x = \frac{- 16 \pm \sqrt{256 - 8 \cdot - 10}}{2 \cdot 2}$

Multiply $- 8$ by $- 10$ .

$x = \frac{- 16 \pm \sqrt{256 + 80}}{2 \cdot 2}$

Add $256$ and $80$ .

$x = \frac{- 16 \pm \sqrt{336}}{2 \cdot 2}$

Rewrite $336$ as $4^{2} \cdot 21$ .

Factor $16$ out of $336$ .

$x = \frac{- 16 \pm \sqrt{16 (21)}}{2 \cdot 2}$

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

$x = \frac{- 16 \pm \sqrt{4^{2} \cdot 21}}{2 \cdot 2}$

Pull terms out from under the radical.

$x = \frac{- 16 \pm 4 \sqrt{21}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{- 16 \pm 4 \sqrt{21}}{4}$

Simplify $\frac{- 16 \pm 4 \sqrt{21}}{4}$ .

$x = - 4 \pm \sqrt{21}$

Step 4

The final answer is the combination of both solutions.

$x = - 4 + \sqrt{21}, - 4 - \sqrt{21}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = - 4 + \sqrt{21}, - 4 - \sqrt{21}$

Decimal Form:

$x = 0.58257569 \dots, - 8.58257569 \dots$