Solve for x 2x^2-4x+5=-3

$2 x^{2} - 4 x + 5 = - 3$

Add $3$ to both sides of the equation.

$2 x^{2} - 4 x + 5 + 3 = 0$

Add $5$ and $3$ .

$2 x^{2} - 4 x + 8 = 0$

Factor $2$ out of $2 x^{2} - 4 x + 8$ .

Factor $2$ out of $2 x^{2}$ .

$2 (x^{2}) - 4 x + 8 = 0$

Factor $2$ out of $- 4 x$ .

$2 (x^{2}) + 2 (- 2 x) + 8 = 0$

Factor $2$ out of $8$ .

$2 x^{2} + 2 (- 2 x) + 2 \cdot 4 = 0$

Factor $2$ out of $2 x^{2} + 2 (- 2 x)$ .

$2 (x^{2} - 2 x) + 2 \cdot 4 = 0$

Factor $2$ out of $2 (x^{2} - 2 x) + 2 \cdot 4$ .

$2 (x^{2} - 2 x + 4) = 0$

Divide each term in $2 (x^{2} - 2 x + 4) = 0$ by $2$ and simplify.

Divide each term in $2 (x^{2} - 2 x + 4) = 0$ by $2$ .

$\frac{2 (x^{2} - 2 x + 4)}{2} = \frac{0}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 (x^{2} - 2 x + 4)}{2} = \frac{0}{2}$

Divide $x^{2} - 2 x + 4$ by $1$ .

$x^{2} - 2 x + 4 = \frac{0}{2}$

Simplify the right side.

Divide $0$ by $2$ .

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

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 2$ , and $c = 4$ into the quadratic formula and solve for $x$ .

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

Simplify.

Simplify the numerator.

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $4$ .

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

Subtract $16$ from $4$ .

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

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

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

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

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

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

$x = \frac{2 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

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

Factor $4$ out of $12$ .

$x = \frac{2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 1}$

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

$x = \frac{2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{2 \pm i \cdot (2 \sqrt{3})}{2 \cdot 1}$

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

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

Multiply $2$ by $1$ .

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

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

$x = 1 \pm i \sqrt{3}$

The final answer is the combination of both solutions.

$x = 1 + i \sqrt{3}, 1 - i \sqrt{3}$

Solve for x 2x^2-4x+5=-3

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