# how many possibilities?number of solutions of equation x^2-square root3*x-6=0

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You need to solve for x the given quadratic equation, hence, you may solve the equation by completing the square `x^2 - sqrt 3*x` such that:

`x^2 - sqrt 3*x + ((sqrt 3)/2)^2 = 6 + ((sqrt 3)/2)^2`

`(x - (sqrt 3)/2)^2 = 27/4 => x - (sqrt 3)/2 = +-sqrt(27/4)`

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

`x_(1,2) = (sqrt 3)/2 +-(3sqrt 3)/2 => {(x_1 = 2sqrt3),(x_2 = -sqrt3):}`

**Hence, evaluating the two real solutions to quadratic equation, yields **`x_1 = 2sqrt3, x_2 = -sqrt3.`

The number of solutions is equal to the maximum superscript of the variable x.

In this case, the maximum number of solutions cannot be greater than 2.

Now, we'll decide if we have 2 different solutions, 2 equal solutions or 2 complex solutions.

The character of solutions is given by the discriminant of the quadratic.

delta = b^2 - 4ac

a,b,c, are the coefficients of the given quadratic.

delta = 3 - 4*1*(-6)

delta = 3 + 24

delta = 27 > 0

Since delta is positive, the quadratic has 2 real unequal solutions.