Answer:
The zeros for this function are x = -1 and x = 1.67
Step-by-step explanation:
Solving a quadratic equation:
Given a second order polynomial expressed by the following equation:
[tex]ax^{2} + bx + c, a\neq0[/tex].
This polynomial has roots [tex]x_{1}, x_{2}[/tex] such that [tex]ax^{2} + bx + c = (x - x_{1})*(x - x_{2})[/tex], given by the following formulas:
[tex]x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}[/tex]
[tex]x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}[/tex]
[tex]\bigtriangleup = b^{2} - 4ac[/tex]
In this question:
[tex]f(x) = 3x^{2} - 2x - 5[/tex]
The zeros of the function are the values of x for which
[tex]f(x) = 0[/tex]
Then
[tex]3x^{2} - 2x - 5 = 0[/tex]
This means that [tex]a = 3, b = -2, c = -5[/tex]
Then
[tex]\bigtriangleup = (-2)^{2} - 4*3*(-5) = 64[/tex]
[tex]x_{1} = \frac{-(-2) + \sqrt{64}}{2*3} = 1.67[/tex]
[tex]x_{2} = \frac{-(-2) - \sqrt{64}}{2*3} = -1[/tex]
The zeros for this function are x = -1 and x = 1.67
