Find All Complex Solutions 5x^2+6x-4=0

$5 x^{2} + 6 x - 4 = 0$

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 $a = 5$ , $b = 6$ , and $c = - 4$ into the quadratic formula and solve for $x$ .

$\frac{- 6 \pm \sqrt{6^{2} - 4 \cdot (5 \cdot - 4)}}{2 \cdot 5}$

Step 3

Simplify.

Simplify the numerator.

Raise $6$ to the power of $2$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 5 \cdot - 4}}{2 \cdot 5}$

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

Multiply $- 4$ by $5$ .

$x = \frac{- 6 \pm \sqrt{36 - 20 \cdot - 4}}{2 \cdot 5}$

Multiply $- 20$ by $- 4$ .

$x = \frac{- 6 \pm \sqrt{36 + 80}}{2 \cdot 5}$

Add $36$ and $80$ .

$x = \frac{- 6 \pm \sqrt{116}}{2 \cdot 5}$

Rewrite $116$ as $2^{2} \cdot 29$ .

Factor $4$ out of $116$ .

$x = \frac{- 6 \pm \sqrt{4 (29)}}{2 \cdot 5}$

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

$x = \frac{- 6 \pm \sqrt{2^{2} \cdot 29}}{2 \cdot 5}$

Pull terms out from under the radical.

$x = \frac{- 6 \pm 2 \sqrt{29}}{2 \cdot 5}$

Multiply $2$ by $5$ .

$x = \frac{- 6 \pm 2 \sqrt{29}}{10}$

Simplify $\frac{- 6 \pm 2 \sqrt{29}}{10}$ .

$x = \frac{- 3 \pm \sqrt{29}}{5}$

Step 4

The final answer is the combination of both solutions.

$x = - \frac{3 - \sqrt{29}}{5}, - \frac{3 + \sqrt{29}}{5}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{3 - \sqrt{29}}{5}, - \frac{3 + \sqrt{29}}{5}$

Decimal Form:

$x = 0.47703296 \dots, - 1.67703296 \dots$