#### Question

In the following, determine whether the given quadratic equation have real roots and if so, find the roots:

`2x^2+5sqrt3x+6=0`

#### Solution

We have been given, `2x^2+5sqrt3x+6=0`

Now we also know that for an equation ax^{2} + bx + c = 0, the discriminant is given by the following equation:

D = b^{2} - 4ac

Now, according to the equation given to us, we have,a = 2, `b=5sqrt3` and c = 6.

Therefore, the discriminant is given as,

`D=(5sqrt3)^2-4(2)(6)`

= 75 - 48

= 27

Since, in order for a quadratic equation to have real roots, D ≥ 0.Here we find that the equation satisfies this condition, hence it has real roots.

Now, the roots of an equation is given by the following equation,

`x=(-b+-sqrtD)/(2a)`

Therefore, the roots of the equation are given as follows,

`x=(-(5sqrt3)+-sqrt27)/(2(2))`

`=(-5sqrt3+-3sqrt3)/4`

Now we solve both cases for the two values of x. So, we have,

`x=(-5sqrt3+3sqrt3)/4`

`=(-sqrt3)/2`

Also,

`x=(-5sqrt3-3sqrt3)/4`

`=-2sqrt3`

Therefore, the roots of the equation are `(-sqrt3)/2` and `-2sqrt3`.