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

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

Step 1

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

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

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

Add $7$ to both sides of the equation.

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

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 $a = 2$ , $b = - 4$ , and $c = 7$ into the quadratic formula and solve for $x$ .

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

Step 4

Simplify.

Simplify the numerator.

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

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

Multiply $- 4 \cdot 2 \cdot 7$ .

Multiply $- 4$ by $2$ .

$x = \frac{4 \pm \sqrt{16 - 8 \cdot 7}}{2 \cdot 2}$

Multiply $- 8$ by $7$ .

$x = \frac{4 \pm \sqrt{16 - 56}}{2 \cdot 2}$

Subtract $56$ from $16$ .

$x = \frac{4 \pm \sqrt{- 40}}{2 \cdot 2}$

Rewrite $- 40$ as $- 1 (40)$ .

$x = \frac{4 \pm \sqrt{- 1 \cdot 40}}{2 \cdot 2}$

Rewrite $\sqrt{- 1 (40)}$ as $\sqrt{- 1} \cdot \sqrt{40}$ .

$x = \frac{4 \pm \sqrt{- 1} \cdot \sqrt{40}}{2 \cdot 2}$

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

$x = \frac{4 \pm i \cdot \sqrt{40}}{2 \cdot 2}$

Rewrite $40$ as $2^{2} \cdot 10$ .

Factor $4$ out of $40$ .

$x = \frac{4 \pm i \cdot \sqrt{4 (10)}}{2 \cdot 2}$

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

$x = \frac{4 \pm i \cdot \sqrt{2^{2} \cdot 10}}{2 \cdot 2}$

Pull terms out from under the radical.

$x = \frac{4 \pm i \cdot (2 \sqrt{10})}{2 \cdot 2}$

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

$x = \frac{4 \pm 2 i \sqrt{10}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{4 \pm 2 i \sqrt{10}}{4}$

Simplify $\frac{4 \pm 2 i \sqrt{10}}{4}$ .

$x = \frac{2 \pm i \sqrt{10}}{2}$