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

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

Step 1

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

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

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

Subtract $2 x$ from both sides of the equation.

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

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 $a = - 3$ , $b = - 2$ , and $c = 1$ into the quadratic formula and solve for $x$ .

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

Step 4

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $- 3$ .

$x = \frac{2 \pm \sqrt{4 + 12 \cdot 1}}{2 \cdot - 3}$

Multiply $12$ by $1$ .

$x = \frac{2 \pm \sqrt{4 + 12}}{2 \cdot - 3}$

Add $4$ and $12$ .

$x = \frac{2 \pm \sqrt{16}}{2 \cdot - 3}$

Rewrite $16$ as $4^{2}$ .

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

Pull terms out from under the radical, assuming positive real numbers.

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

Multiply $2$ by $- 3$ .

$x = \frac{2 \pm 4}{- 6}$

Simplify $\frac{2 \pm 4}{- 6}$ .

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

Move the negative in front of the fraction.

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

Step 5

The final answer is the combination of both solutions.

$x = - 1, \frac{1}{3}$