Question

x^{2}+(1 + sqrt{3})x-2-sqrt{3}=0

Explanation:

Step1: Identify coefficients for quadratic formula

For the quadratic equation $x^{2}+(1 + \sqrt{3})x-2-\sqrt{3}=0$, we have $a = 1$, $b=1+\sqrt{3}$, $c=-2 - \sqrt{3}$.

Step2: Calculate the discriminant $\Delta$

$\Delta=b^{2}-4ac=(1 + \sqrt{3})^{2}-4\times1\times(-2-\sqrt{3})$.
First expand $(1 + \sqrt{3})^{2}=1 + 2\sqrt{3}+3=4 + 2\sqrt{3}$.
Then $4\times1\times(-2-\sqrt{3})=-8 - 4\sqrt{3}$.
So $\Delta=4 + 2\sqrt{3}+8 + 4\sqrt{3}=12 + 6\sqrt{3}=3(4 + 2\sqrt{3})=3(1+\sqrt{3})^{2}$.

Step3: Use the quadratic formula $x=\frac{-b\pm\sqrt{\Delta}}{2a}$

$x=\frac{-(1 + \sqrt{3})\pm\sqrt{3(1+\sqrt{3})^{2}}}{2\times1}=\frac{-(1 + \sqrt{3})\pm(1+\sqrt{3})\sqrt{3}}{2}$.
Case 1: $x=\frac{-(1 + \sqrt{3})+(1+\sqrt{3})\sqrt{3}}{2}=\frac{-1-\sqrt{3}+\sqrt{3}+3}{2}=1$.
Case 2: $x=\frac{-(1 + \sqrt{3})-(1+\sqrt{3})\sqrt{3}}{2}=\frac{-1-\sqrt{3}-\sqrt{3}-3}{2}=-2-\sqrt{3}$.

Answer:

$x = 1$ or $x=-2-\sqrt{3}$