Answer:
x = [tex]\frac{12+\sqrt{138} }{6}[/tex] and x = [tex]\frac{12-\sqrt{138} }{6}[/tex]
Step-by-step explanation:
Let's use the quadratic formula, which states that for a quadratic of the form ax² + bc + c, the zeroes are: [tex]x=\frac{-b+\sqrt{b^2-4ac} }{2a}[/tex] or [tex]x=\frac{-b-\sqrt{b^2-4ac} }{2a}[/tex].
Here, a = 6, b = -24, and c = 1. Plug these in:
[tex]x=\frac{-b+\sqrt{b^2-4ac} }{2a}[/tex]
[tex]x=\frac{-(-24)+\sqrt{(-24)^2-4*6*1} }{2*6}=\frac{24+\sqrt{552} }{12}=\frac{24+2\sqrt{138} }{12} =\frac{12+\sqrt{138} }{6}[/tex]
AND
[tex]x=\frac{-b-\sqrt{b^2-4ac} }{2a}[/tex]
[tex]x=\frac{-(-24)-\sqrt{(-24)^2-4*6*1} }{2*6}=\frac{24-\sqrt{552} }{12}=\frac{24-2\sqrt{138} }{12} =\frac{12-\sqrt{138} }{6}[/tex]
Thus, the zeroes are: x = [tex]\frac{12+\sqrt{138} }{6}[/tex] and x = [tex]\frac{12-\sqrt{138} }{6}[/tex].
