Find the Roots (Zeros) f(x)=6x^2+10x-1

$f (x) = 6 x^{2} + 10 x - 1$

Step 1

Set $6 x^{2} + 10 x - 1$ equal to $0$ .

$6 x^{2} + 10 x - 1 = 0$

Step 2

Solve for $x$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 6$ , $b = 10$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

$\frac{- 10 \pm \sqrt{10^{2} - 4 \cdot (6 \cdot - 1)}}{2 \cdot 6}$

Simplify.

Simplify the numerator.

Raise $10$ to the power of $2$ .

$x = \frac{- 10 \pm \sqrt{100 - 4 \cdot 6 \cdot - 1}}{2 \cdot 6}$

Multiply $- 4 \cdot 6 \cdot - 1$ .

Multiply $- 4$ by $6$ .

$x = \frac{- 10 \pm \sqrt{100 - 24 \cdot - 1}}{2 \cdot 6}$

Multiply $- 24$ by $- 1$ .

$x = \frac{- 10 \pm \sqrt{100 + 24}}{2 \cdot 6}$

Add $100$ and $24$ .

$x = \frac{- 10 \pm \sqrt{124}}{2 \cdot 6}$

Rewrite $124$ as $2^{2} \cdot 31$ .

Factor $4$ out of $124$ .

$x = \frac{- 10 \pm \sqrt{4 (31)}}{2 \cdot 6}$

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

$x = \frac{- 10 \pm \sqrt{2^{2} \cdot 31}}{2 \cdot 6}$

Pull terms out from under the radical.

$x = \frac{- 10 \pm 2 \sqrt{31}}{2 \cdot 6}$

Multiply $2$ by $6$ .

$x = \frac{- 10 \pm 2 \sqrt{31}}{12}$

Simplify $\frac{- 10 \pm 2 \sqrt{31}}{12}$ .

$x = \frac{- 5 \pm \sqrt{31}}{6}$

The final answer is the combination of both solutions.

$x = - \frac{5 - \sqrt{31}}{6}, - \frac{5 + \sqrt{31}}{6}$

$x = \frac{- 5 \pm \sqrt{31}}{6}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = \frac{- 5 \pm \sqrt{31}}{6}$

Decimal Form:

$x = 0.09462739 \dots, - 1.76129406 \dots$

Step 4