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

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

Move all terms to the left side of the equation and simplify.

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

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

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

Add $3$ to both sides of the equation.

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

Add $1$ and $3$ .

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

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

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