Answer:
r = 6, r = - ⅔
Step-by-step explanation:
Given the quadratic equation, 3r² - 16r - 12 = 0:
where a = 3, b = -16, and c = -12.
Use the quadratic formula to find the x-intercepts:
[tex]r = \frac{-b +/- \sqrt{b^{2}-4ac } }{2a}[/tex]
[tex]r = \frac{16 +/- \sqrt{(-16)^{2}-4(3)(-12) } }{2(3)}[/tex]
[tex]r = \frac{16 +/- \sqrt{256 - 144 } }{6}[/tex]
[tex]r = \frac{16 +/- \sqrt{400} }{6}[/tex]
[tex]r = \frac{16 + 20}{6} = \frac{36}{6} = 6[/tex] , [tex]r = \frac{16 - 20}{6} = - \frac{4}{6} = -\frac{2}{3}[/tex]
Therefore, the roots (zeroes) of the given quadratic equation are:
r = 6, r = - ⅔
