How do you solve #6h^2+17h+12=0#?

# 6h^2 + 17h +12 = 0#

We can Split the Middle Term of this equation to factorise it and then find the solution for the equation

In this technique, if we have to factorise an expression like #ah^2 + bh + c#, we need to think of 2 numbers such that:
#N_1 xx N_2 = a xx c = 6 xx 12 = 72#
and
#N_1 +N_2 = b = 17#

After trying out a few numbers we get #N_1 = 9# and #N_2 =8#
#9 xx 8 = 72#, and # 9 +8 = 17#

# 6h^2 + 17h +12 = 6h^2 + 9h +8h +12 #
# = 3h (2h + 3) + 4(2h + 3) #
# = (3h +4) (2h + 3) #
the solution for the equation is :
#color(blue)h=-4/3 #
#color(blue)h= -3/2#

The answer is #h=-3/2, -4/3#

Problem: Solve #6h^2+17h+12=0# .

Multiply #6# times #12# to get #72#.
Find two numbers that when multiplied equal #72# and when added equal #17#.
Numbers #8# and #9# meet the criteria.
Write the equation substituting #8h# and #9h# for #17h#.

#6h^2+8h+9h+12=0#

Separate the first two terms from the second two terms.

#(6h^2+8h)+(9h+12)#

Factor out the GCF from both sets of terms.

#2h(3h+4)+3(3h+4)=0#

Factor out the GCF #(3h+4)# .

#(3h+4)(2h+3)=0#

Solve for #h#.

#3h+4=0#

#3h=-4#

#h=-4/3#

#2h+3=0#

#2h=-3#

#h=-3/2#

#h=-3/2, -4/3#