Solve for x 6-6x^2=3x

$6 - 6 x^{2} = 3 x$

Subtract $3 x$ from both sides of the equation.

$6 - 6 x^{2} - 3 x = 0$

Factor $- 3$ out of $6 - 6 x^{2} - 3 x$ .

Move $6$ .

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

Factor $- 3$ out of $- 6 x^{2}$ .

$- 3 (2 x^{2}) - 3 x + 6 = 0$

Factor $- 3$ out of $- 3 x$ .

$- 3 (2 x^{2}) - 3 x + 6 = 0$

Factor $- 3$ out of $6$ .

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

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

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

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

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

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

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

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

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

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

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

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

Simplify the right side.

Divide $0$ by $- 3$ .

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

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 2$ , $b = 1$ , and $c = - 2$ into the quadratic formula and solve for $x$ .

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

Simplify.

Simplify the numerator.

One to any power is one.

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

Multiply $- 4$ by $2$ .

$x = \frac{- 1 \pm \sqrt{1 - 8 \cdot - 2}}{2 \cdot 2}$

Multiply $- 8$ by $- 2$ .

$x = \frac{- 1 \pm \sqrt{1 + 16}}{2 \cdot 2}$

Add $1$ and $16$ .

$x = \frac{- 1 \pm \sqrt{17}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{- 1 \pm \sqrt{17}}{4}$

The final answer is the combination of both solutions.

$x = - \frac{1 - \sqrt{17}}{4}, - \frac{1 + \sqrt{17}}{4}$

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{1 - \sqrt{17}}{4}, - \frac{1 + \sqrt{17}}{4}$

Decimal Form:

$x = 0.78077640 \dots, - 1.28077640 \dots$

Solve for x 6-6x^2=3x

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