(a)
#2x^2+11x+15=0#
#=(2x \ \ \ \ \ \ \ )( x \ \ \ \ \ \ \ )=0# with gaps to be filled
#=(2x + 5)(x+3)=0# guessing that the 5 and 3 go in that order and checking, or guessing wrong first!
Hence
#x=-5/2# or #x=-3#
(b) Quadratic formula is that the equation #ax²+bx+c=0# has solutions #(-b±sqrt(b²-4ac))/(2a)#. Here #a=2#, #b=12#, #c=18#.
So the solutions are #x=(-12±sqrt(12² -4×2×18))/(2×2)#
# = (-12±sqrt(0))/4#
#=-3# (the case of "two equal roots")
(Or you could have factorized as #2(x²+6x+9)=2(x+3)²#)
(c)
#x²+10x+24=0#
#(x+5)²-25+24=0# (Halve the 10 to get the 5, then subtract the 25)
#(x+5)²-1=0#
#(x+5)²=+1#
#x+5=±1# (Don't forget the ±!)
#x=-5-1# or #-5+1#
#x=-6# or #-4#
Then check all solutions by substituting the values of #x# back into the original equations.
