#6x^2+3x-63 = (3*2x^2)+(3*x)-(3*21)#
#= 3(2x^2+x-21)#
To factor #2x^2+x-21# I will look for roots of the equation:
#2x^2+x-21 = 0#
This is of the form #ax^2+bx+c = 0#, which has solutions:
#x=(-b+-sqrt(b^2-4ac))/(2a)#
In our case #a=2#, #b=1# and #c=-21#, so
#x=(-b+-sqrt(b^2-4ac))/(2a)#
#=(-1+-sqrt(1^2-4*2*(-21)))/(2*2)#
#=(-1+-sqrt(1+168))/4#
#=(-1+-sqrt(169))/4#
#=(-1+-13)/4#
So one root is #12/4 = 3# and the other is #-14/4 = -7/2#.
From this we can deduce that #(x-3)# is a factor and #(2x+7)# is the other factor.
In summary, we have #6x^2+3x-63 = 3(x-3)(2x+7)#.
