Solve for x -6+3x-9x^2=-16

$- 6 + 3 x - 9 x^{2} = - 16$

Move all terms to the left side of the equation and simplify.

Add $16$ to both sides of the equation.

$- 6 + 3 x - 9 x^{2} + 16 = 0$

Add $- 6$ and $16$ .

$3 x - 9 x^{2} + 10 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = - 9$ , $b = 3$ , and $c = 10$ into the quadratic formula and solve for $x$ .

$\frac{- 3 \pm \sqrt{3^{2} - 4 \cdot (- 9 \cdot 10)}}{2 \cdot - 9}$

Simplify.

Simplify the numerator.

Raise $3$ to the power of $2$ .

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot - 9 \cdot 10}}{2 \cdot - 9}$

Multiply $- 4$ by $- 9$ .

$x = \frac{- 3 \pm \sqrt{9 + 36 \cdot 10}}{2 \cdot - 9}$

Multiply $36$ by $10$ .

$x = \frac{- 3 \pm \sqrt{9 + 360}}{2 \cdot - 9}$

Add $9$ and $360$ .

$x = \frac{- 3 \pm \sqrt{369}}{2 \cdot - 9}$

Rewrite $369$ as $3^{2} \cdot 41$ .

Factor $9$ out of $369$ .

$x = \frac{- 3 \pm \sqrt{9 (41)}}{2 \cdot - 9}$

Rewrite $9$ as $3^{2}$ .

$x = \frac{- 3 \pm \sqrt{3^{2} \cdot 41}}{2 \cdot - 9}$

Pull terms out from under the radical.

$x = \frac{- 3 \pm 3 \sqrt{41}}{2 \cdot - 9}$

Multiply $2$ by $- 9$ .

$x = \frac{- 3 \pm 3 \sqrt{41}}{- 18}$

Simplify $\frac{- 3 \pm 3 \sqrt{41}}{- 18}$ .

$x = \frac{1 \pm \sqrt{41}}{6}$

The final answer is the combination of both solutions.

$x = \frac{1 + \sqrt{41}}{6}, \frac{1 - \sqrt{41}}{6}$

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1 + \sqrt{41}}{6}, \frac{1 - \sqrt{41}}{6}$

Decimal Form:

$x = 1.23385403 \dots, - 0.90052070 \dots$