Find the Roots (Zeros) 6x^2+2x-18=0

$6 x^{2} + 2 x - 18 = 0$

Factor $2$ out of $6 x^{2} + 2 x - 18$ .

Factor $2$ out of $6 x^{2}$ .

$2 (3 x^{2}) + 2 x - 18 = 0$

Factor $2$ out of $2 x$ .

$2 (3 x^{2}) + 2 (x) - 18 = 0$

Factor $2$ out of $- 18$ .

$2 (3 x^{2}) + 2 x + 2 \cdot - 9 = 0$

Factor $2$ out of $2 (3 x^{2}) + 2 x$ .

$2 (3 x^{2} + x) + 2 \cdot - 9 = 0$

Factor $2$ out of $2 (3 x^{2} + x) + 2 \cdot - 9$ .

$2 (3 x^{2} + x - 9) = 0$

Divide each term in $2 (3 x^{2} + x - 9) = 0$ by $2$ and simplify.

Divide each term in $2 (3 x^{2} + x - 9) = 0$ by $2$ .

$\frac{2 (3 x^{2} + x - 9)}{2} = \frac{0}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 (3 x^{2} + x - 9)}{2} = \frac{0}{2}$

Divide $3 x^{2} + x - 9$ by $1$ .

$3 x^{2} + x - 9 = \frac{0}{2}$

Simplify the right side.

Divide $0$ by $2$ .

$3 x^{2} + x - 9 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 3$ , $b = 1$ , and $c = - 9$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (3 \cdot - 9)}}{2 \cdot 3}$

Simplify.

Simplify the numerator.

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 3 \cdot - 9}}{2 \cdot 3}$

Multiply $- 4$ by $3$ .

$x = \frac{- 1 \pm \sqrt{1 - 12 \cdot - 9}}{2 \cdot 3}$

Multiply $- 12$ by $- 9$ .

$x = \frac{- 1 \pm \sqrt{1 + 108}}{2 \cdot 3}$

Add $1$ and $108$ .

$x = \frac{- 1 \pm \sqrt{109}}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{- 1 \pm \sqrt{109}}{6}$

The result can be shown in multiple forms.

Exact Form:

$x = \frac{- 1 \pm \sqrt{109}}{6}$

Decimal Form:

$x = 1.57338441 \dots, - 1.90671775 \dots$