Answer:
[tex]f(x) = 6x ^{2} + 12x - 7 \\ x = \frac{ - 6 + \sqrt{78} }{6} \: , - \frac{ - 6 - \sqrt{78} }{6} [/tex]
Given:
The quadratic function is:
[tex]f(x)=6x^2+12x-7[/tex]
To find:
The zeros of the quadratic function.
Solution:
Quadratic formula: If a quadratic equation is [tex]ax^2+bx+c=0[/tex], then zeros of the quadratic equation are:
[tex]x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
We have,
[tex]f(x)=6x^2+12x-7[/tex]
For zeros, [tex]f(x)=0[/tex].
[tex]6x^2+12x-7=0[/tex]
Here, [tex]a=6,b=12,c=-7[/tex]. Using quadratic formula, we get
[tex]x=\dfrac{-12\pm \sqrt{(12)^2-4(6)(-7)}}{2(6)}[/tex]
[tex]x=\dfrac{-12\pm \sqrt{144+168}}{12}[/tex]
[tex]x=\dfrac{-12\pm \sqrt{312}}{12}[/tex]
[tex]x=\dfrac{-12\pm 17.6635}{12}[/tex]
Now,
[tex]x=\dfrac{-12+17.6635}{12}[/tex] and [tex]x=\dfrac{-12-17.6635}{12}[/tex]
[tex]x=0.47195833...[/tex] and [tex]x=-2.47195833...[/tex]
[tex]x\approx 0.472[/tex] and [tex]x\approx -2.472[/tex]
Therefore, the zeros of the given quadratic function are 0.472 and -2.472.
