How do you solve #2x^2-5x=4#?

1 Answer
Alan P.
May 30, 2015

#2x^2-5x=4#
is equivalent to
#color(white)("XXXXX")##2x^2-5x-4=0#

For a quadratic in the general form
#color(white)("XXXXX")##ax^2+bx+c = 0#
the solutions can be obtained using the quadratic root formula
#color(white)("XXXXX")##x=(-b+-sqrt(b^2-4ac))/(2a)#

In this case the roots are:
#color(white)("XXXXX")##x = (5+-sqrt(25+32))/4#

#x= (5+sqrt(57))/4# or #x=(5-sqrt(57))/4#

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