Solve Using the Quadratic Formula x^2-4x=-3

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

Step 1

Add $3$ to both sides of the equation.

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

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 $a = 1$ , $b = - 4$ , and $c = 3$ into the quadratic formula and solve for $x$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (1 \cdot 3)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

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

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot 3}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 3$ .

Multiply $- 4$ by $1$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 3}}{2 \cdot 1}$

Multiply $- 4$ by $3$ .

$x = \frac{4 \pm \sqrt{16 - 12}}{2 \cdot 1}$

Subtract $12$ from $16$ .

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

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

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

Pull terms out from under the radical, assuming positive real numbers.

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

Multiply $2$ by $1$ .

$x = \frac{4 \pm 2}{2}$

Simplify $\frac{4 \pm 2}{2}$ .

$x = 2 \pm 1$

Step 5

The final answer is the combination of both solutions.

$x = 3, 1$