Solve for x x^2-4x-3=0

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

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

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 3

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 $- 3$ .

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

Add $16$ and $12$ .

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

Rewrite $28$ as $2^{2} \cdot 7$ .

Factor $4$ out of $28$ .

$x = \frac{4 \pm \sqrt{4 (7)}}{2 \cdot 1}$

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

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

Pull terms out from under the radical.

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

Multiply $2$ by $1$ .

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

Simplify $\frac{4 \pm 2 \sqrt{7}}{2}$ .

$x = 2 \pm \sqrt{7}$

Step 4

The final answer is the combination of both solutions.

$x = 2 + \sqrt{7}, 2 - \sqrt{7}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = 2 + \sqrt{7}, 2 - \sqrt{7}$

Decimal Form:

$x = 4.64575131 \dots, - 0.64575131 \dots$