How do you solve #3x^2+9x-12=0# using the quadratic formula?

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Answer 1 · 2016-02-15T06:11:26

Solution is #x=1 or -4#.

Solution of quadratic equation #ax^2+bx+c=0# is given by

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

Here #a=3, b=9 and c=-12#. Putting these values

#x=(-9+-sqrt(9^2-4*3*(-12)))/(2*3)#

= #x=(-9+-sqrt(81+144))/6#

= #x=(-9+-sqrt(225))/6# = #x=(-9+-15)/6#

= #1 or -4#