Solve Using the Quadratic Formula 5x^2+6x-1=0

$5 x^{2} + 6 x - 1 = 0$

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 $c = - 1$ into the quadratic formula and solve for $x$ .

$\frac{- 6 \pm \sqrt{6^{2} - 4 \cdot (5 \cdot - 1)}}{2 \cdot 5}$

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 5 \cdot - 1}}{2 \cdot 5}$

Multiply $- 4 \cdot 5 \cdot - 1$ .

Multiply $- 4$ by $5$ .

$x = \frac{- 6 \pm \sqrt{36 - 20 \cdot - 1}}{2 \cdot 5}$

Multiply $- 20$ by $- 1$ .

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

Add $36$ and $20$ .

$x = \frac{- 6 \pm \sqrt{56}}{2 \cdot 5}$

Rewrite $56$ as $2^{2} \cdot 14$ .

Factor $4$ out of $56$ .

$x = \frac{- 6 \pm \sqrt{4 (14)}}{2 \cdot 5}$

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

$x = \frac{- 6 \pm \sqrt{2^{2} \cdot 14}}{2 \cdot 5}$

Pull terms out from under the radical.

$x = \frac{- 6 \pm 2 \sqrt{14}}{2 \cdot 5}$

Multiply $2$ by $5$ .

$x = \frac{- 6 \pm 2 \sqrt{14}}{10}$

Simplify $\frac{- 6 \pm 2 \sqrt{14}}{10}$ .

$x = \frac{- 3 \pm \sqrt{14}}{5}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \frac{- 3 \pm \sqrt{14}}{5}$

Decimal Form:

$x = 0.14833147 \dots, - 1.34833147 \dots$