Question #de83c

Question

973 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-02T13:23:57

#x=5# or #x=(-5)/2#

#2x^2-25-5x=0#
#2x^2-5x-25=0#
#2x^2-10x+5x-25=0#
#2x(x-5)+5(x-5)=0#
#(2x+5)(x-5)=0#
#(2x+5)=0# or #x-5=0#
#2x=-5# or #x=5#
#x=(-5)/2# or #x=5#

Answer 2 · 2017-11-02T14:12:56

#x=5,-5/2#

The quadratic formula is #x=(-b+-sqrt(b^2-4ac))/(2a)#

Given equation is #2x^2-5x-25#

Here #a=2# , #b=-5# and #c=-25#

Put these values in the quadratic formula.

#x=(-(-5)+-sqrt((-5)^2-4 xx (2) xx (-25)))/(2 xx 2)#

#x=(5+-sqrt(25+200))/4#

#x=(5+-15)/4#

So #x=(5+15)/4# or #x=(5-15)/4#

#x=20/4# or #x=(-10)/4#

#x=5# or #x=(-5)/2#