How do you solve #5x^2 + x = 3# using the quadratic formula?

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Answer 1 · 2016-03-24T01:51:11

#x = (-1 +- sqrt61)/10#

Given

#ax^2 + bx + c = 0#

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

#5x^2 + x = 3 => 5x^2 + x - 3 = 0#

#=> a = 5#
#=> b = 1#
#=> c = -3#

#x = (-1 +- sqrt(1^2 - 4*5*-3))/(2*5)#

#x = (-1 +- sqrt(1 + 60))/(10)#

#x = (-1 +- sqrt61)/10#