#### Question

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

`3x^2+2sqrt5x-5=0`

#### Solution

We have been given, `3x^2+2sqrt5x-5=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 = 3, `b=2sqrt5` and c = -5.

Therefore, the discriminant is given as,

`D=(2sqrt5)^2-4(3)(-5)`

= 20 + 60

= 80

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=(-(2sqrt5)+-sqrt80)/(2(3))`

`=(-2sqrt5+-4sqrt5)(2(3))`

`=(-sqrt5+-2sqrt5)/3`

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

`x=(-sqrt5+2sqrt5)/3`

`=sqrt5/3`

Also,

`=(-sqrt5-2sqrt5)/3`

`=-sqrt5`

Therefore, the roots of the equation are `sqrt5/3` and `-sqrt5`.