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

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

Step 1

Subtract $1$ from both sides of the equation.

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

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 $a = 3$ , $b = 2$ , 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 $\frac{- 2 \pm 4}{6}$ .

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

Step 5

The final answer is the combination of both solutions.

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