Answer:
The roots are [tex]x = (\frac{3}{2}, -2)[/tex], which is given by option C.
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 = a(x - x_{1})*(x - x_{2})[/tex], given by the following formulas:
[tex]x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}[/tex]
[tex]x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}[/tex]
[tex]\Delta = b^{2} - 4ac[/tex]
In this question, we have that:
[tex]2x^2 + x - 6 = 0[/tex]
Which is a quadratic equation with [tex]a = 2, b = 1, c = -6[/tex]. So
[tex]\Delta = 1^{2} - 4*2(-6) = 1 + 48 = 49[/tex]
[tex]x_{1} = \frac{-1 + \sqrt{49}}{2*2} = \frac{6}{4} = \frac{3}{2}[/tex]
[tex]x_{2} = \frac{-1 - \sqrt{49}}{2*2} = \frac{-8}{4} = -2[/tex]
So the roots are [tex]x = (\frac{3}{2}, -2)[/tex], which is given by option C.
