Answer
$\left\{\dfrac{-\sqrt{3}- \sqrt{15}}{2},\dfrac{-\sqrt{3}+ \sqrt{15}}{2}\right\}$.
Work Step by Step
The equation is in standard form $ax^2+bx+c=0$.
We have $a=1,b=\sqrt{3}$ and $c=-3$.
The discriminant is
$=b^2-4ac$
$=(\sqrt{3})^2-4(1)(-3)$
$=3+12$
$=15$
Since $15>0$. There are two real solutions.
The quadratic formula is
$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$
Substitute the values of $a, b, c, $ and the discriminant to obtain:.
$x=\dfrac{-(\sqrt{3})\pm \sqrt{(15)}}{2(1)}$
Simplify.
$x=\dfrac{-\sqrt{3}\pm \sqrt{15}}{2}$
Hence, the solution set is $\left\{\dfrac{-\sqrt{3}- \sqrt{15}}{2},\dfrac{-\sqrt{3}+ \sqrt{15}}{2}\right\}$.
