How do you solve #4c^2 + 10c = -7# by completing the square?

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Answer 1 · 2018-03-20T21:28:05

#c= -5/4 +- sqrt(-3)/4#

Factor out the coefficient of #x^2#, then work with the quadratic expression which has a coefficient of #x^2# equal to #1#
Check the coefficient of #x# in the new quadratic expression and take half of it - this is the number that goes into the complete square bracket
Balance the constant term by subtracting the square of the number from step 2 and putting in the constant from the quadratic expression
The rest is arithmetic that may often involve fractions

Re-write the equations as

#4c^2+10c+7 = 0 #

#4(c^2 + (10c)/4+7/4) =0#

#4[(c+10/(4*2))^2 - (5/4)^2 + 7/4] =0#

#4[(c+5/4)^2 - 25/16 + 7/4] = 0#

#4[(c+5/4)^2 + 3/16] = 0 #

#4(c+5/4)^2 = - 3/16 * 4#

#2(c+5/4) = +- sqrt( - 3/4)#

#c= -5/4 +- sqrt(-3)/4#

This just presents the quadratic in an alternative format.