How do you solve # 8x^2 + 2x - 3 = 0# using the quadratic formula?

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Answer 1 · 2016-05-04T21:13:52

#x = 1/2# or #x = -3/4#

Given:

#8x^2+2x-3 = 0#

This is in the form #ax^2+bx+c = 0# with #a=8#, #b=2# and #c=-3#.

It has roots given by the quadratic formula:

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

#=(-2+-sqrt(2^2-(4*8*(-3))))/(2*8)#

#=(-2+-sqrt(4+96))/16#

#=(-2+-sqrt(100))/16#

#=(-2+-10)/16#

That is:

#x = (-2+10)/16 = 8/16 = 1/2#

or:

#x = (-2-10)/16 = (-12)/16 = -3/4#