Answer:
The solutions for x are [tex]x = 4[/tex] and [tex]x = -1.5[/tex].
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]
2x² - 5x - 12 = 0
This means that [tex]a = 2, b = -5, c = -12[/tex]
Solution:
[tex]\Delta = (-5)^2 - 4(2)(-12) = 121[/tex]
[tex]x_{1} = \frac{-(-5) + \sqrt{121}}{2*2} = 4[/tex]
[tex]x_{2} = \frac{-(-5) - \sqrt{121}}{2*2} = -1.5[/tex]
The solutions for x are [tex]x = 4[/tex] and [tex]x = -1.5[/tex].
