How do you solve #3x ^ { 2} + 21x = - 36# using the quadratic formula?

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Answer 1 · 2017-03-31T18:54:36

#x = -4, " and " x = -3#

#x =( -B+-sqrt(B^2-4AC))/(2A) # with the form #Ax^2 + Bx + C = 0#

#3x^2 + 21x +36 = 0#

#x = (-21 +- sqrt(441 - 4*3*36))/(2*3) = -21/6 +- sqrt (9)/6 = -7/2+- 3/6#

#x = -7/2 +- 1/2#

#x = -7/2 + 1/2 = -6/2 = -3 " and"#

#x = -7/2 - 1/2 = -8/2 = -4#