Answer:
[tex]x=-1+\frac{\sqrt{78} }{6}[/tex] and
[tex]x=-1-\frac{\sqrt{78} }{6}[/tex]
Step-by-step explanation:
[tex]f(x) = 6x^2 + 12x - 7[/tex]
To find out the zeros of the quadratic function, we apply quadratic formula
[tex]x=\frac{-b+-\sqrt{b^2-4ac}}{2a}[/tex]
From the given f(x), the value of a=6, b=12, c=-7
Plug in all the values in the formula
[tex]x=\frac{-12+-\sqrt{12^2-4(6)(-7)}}{2(6)}[/tex]
[tex]x=\frac{-12+-\sqrt{312}}{2(6)}[/tex]
[tex]x=\frac{-12+-2\sqrt{78}}{2(6)}[/tex]
Now divide each term by 12
[tex]x=-1+-\frac{\sqrt{78} }{6}[/tex]
We will get two values for x
[tex]x=-1+\frac{\sqrt{78} }{6}[/tex] and
[tex]x=-1-\frac{\sqrt{78} }{6}[/tex]
