Solve Using the Quadratic Formula -6x=3x^2+1

$- 6 x = 3 x^{2} + 1$

Step 1

Move all the expressions to the left side of the equation.

Subtract $3 x^{2}$ from both sides of the equation.

$- 6 x - 3 x^{2} = 1$

Subtract $1$ from both sides of the equation.

$- 6 x - 3 x^{2} - 1 = 0$

Step 2

Use the quadratic formula to find the solutions.

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

Step 3

Substitute the values $a = - 3$ , $b = - 6$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (- 3 \cdot - 1)}}{2 \cdot - 3}$

Step 4

Simplify.

Simplify the numerator.

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

$x = \frac{6 \pm \sqrt{36 - 4 \cdot - 3 \cdot - 1}}{2 \cdot - 3}$

Multiply $- 4 \cdot - 3 \cdot - 1$ .

Multiply $- 4$ by $- 3$ .

$x = \frac{6 \pm \sqrt{36 + 12 \cdot - 1}}{2 \cdot - 3}$

Multiply $12$ by $- 1$ .

$x = \frac{6 \pm \sqrt{36 - 12}}{2 \cdot - 3}$

Subtract $12$ from $36$ .

$x = \frac{6 \pm \sqrt{24}}{2 \cdot - 3}$

Rewrite $24$ as $2^{2} \cdot 6$ .

Factor $4$ out of $24$ .

$x = \frac{6 \pm \sqrt{4 (6)}}{2 \cdot - 3}$

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

$x = \frac{6 \pm \sqrt{2^{2} \cdot 6}}{2 \cdot - 3}$

Pull terms out from under the radical.

$x = \frac{6 \pm 2 \sqrt{6}}{2 \cdot - 3}$

Multiply $2$ by $- 3$ .

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

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

$x = \frac{3 \pm \sqrt{6}}{- 3}$

Move the negative in front of the fraction.

$x = - \frac{3 \pm \sqrt{6}}{3}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{3 \pm \sqrt{6}}{3}$

Decimal Form:

$x = - 1.81649658 \dots, - 0.18350341 \dots$