Solve Using the Quadratic Formula 7x^2=4x-1

$7 x^{2} = 4 x - 1$

Step 1

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

Subtract $4 x$ from both sides of the equation.

$7 x^{2} - 4 x = - 1$

Add $1$ to both sides of the equation.

$7 x^{2} - 4 x + 1 = 0$

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 $a = 7$ , $b = - 4$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (7 \cdot 1)}}{2 \cdot 7}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 4$ to the power of $2$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 7 \cdot 1}}{2 \cdot 7}$

Multiply $- 4 \cdot 7 \cdot 1$ .

Multiply $- 4$ by $7$ .

$x = \frac{4 \pm \sqrt{16 - 28 \cdot 1}}{2 \cdot 7}$

Multiply $- 28$ by $1$ .

$x = \frac{4 \pm \sqrt{16 - 28}}{2 \cdot 7}$

Subtract $28$ from $16$ .

$x = \frac{4 \pm \sqrt{- 12}}{2 \cdot 7}$

Rewrite $- 12$ as $- 1 (12)$ .

$x = \frac{4 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 7}$

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$x = \frac{4 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 7}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{4 \pm i \cdot \sqrt{12}}{2 \cdot 7}$

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$x = \frac{4 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 7}$

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

$x = \frac{4 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 7}$

Pull terms out from under the radical.

$x = \frac{4 \pm i \cdot (2 \sqrt{3})}{2 \cdot 7}$

Move $2$ to the left of $i$ .

$x = \frac{4 \pm 2 i \sqrt{3}}{2 \cdot 7}$

Multiply $2$ by $7$ .

$x = \frac{4 \pm 2 i \sqrt{3}}{14}$

Simplify $\frac{4 \pm 2 i \sqrt{3}}{14}$ .

$x = \frac{2 \pm i \sqrt{3}}{7}$