Question

How do you solve `3x ^ { 2} + x - 5= 0`?

Answer 1

The values that x assume are:
`x_1 = (-1 + sqrt (61))/6` and `x_2 = (-1 - sqrt (61))/6`

Explanation:

Through bhaskara's formula:
A second degree equation has the general formula:
`ax^2 +bx +c = 0`

First step, find the delta:
`Delta = b^2 -4ac = 1^2 -4 (3)(-5) = 61`

Delta being found, the values of x are defined by this formula:
`x =( -b +- sqrt (Delta))/(2a)`

So, the values that x assume are:
`x_1 = (-1 + sqrt (61))/6` and `x_2 = (-1 - sqrt (61))/6`