Answer:
A quadratic equation is of the form: \(ax^2+bx+c= 0\) .....[1] where
a. b and c are coefficient and x is the variable.
The solutions of the equation are;
\(x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\)
Given the function: \(f(x) = 6x^2+12x-7\)
To find the zero of this function.
Set f(x) = 0
⇒\(6x^2+12x-7=0\)
On comparing this with equation [1] we have;
a = 6, b = 12 and c = -7
then;
\(x = \frac{-12\pm\sqrt{(12)^2-4(6)(-7)}}{2(6)}\)
\(x = \frac{-12\pm\sqrt{144+168}}{12}\)
or
\(x = \frac{-12\pm\sqrt{312}}{12}\)
or
\(x = \frac{-12\pm 2\sqrt{78}}{12}\)
\(x = \frac{-6\pm \sqrt{78}}{6}\)
Therefore, the zeros of the quadratic function f(x) are;
\(\frac{-6+\sqrt{78}}{6}\) and \(\frac{-6-\sqrt{78}}{6}\)
