What are the zeros of the quadratic function #f(x)=6x^2+12x-7#?

1 Answer
Nov 19, 2016

Answer:

#x = -1+-1/6 sqrt(78)#

Explanation:

#f(x) = 6x^2+12x-7#

is of the form #ax^2+bx+c# with #a=6#, #b=12# and #c=-7#

We can find the zeros of #f(x)# using the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#color(white)(x) = (-12+-sqrt(12^2-4(6)(-7)))/(2(6))#

#color(white)(x) = (-12+-sqrt(144+168))/12#

#color(white)(x) = (-12+-sqrt(312))/12#

#color(white)(x) = (-12+-sqrt(4*78))/12#

#color(white)(x) = (-12+-2sqrt(78))/12#

#color(white)(x) = -1+-1/6 sqrt(78)#