The constant 3 is a prime number so the only factors of 3 are 1 and 3. So our stating point is:
#(?w+1)(?w+3)#
The constant is positive in that we have +3. This means that the signs in the brackets are both the same.
As the signs are the same then the plus from #+15w# means the two signs are both positive. Thus we also have the correct structure in that #(?w+1)(?w+3)# is correct.
Consider the whole number factors of 10 #-> 1xx10" and "2xx5#
#color(red)("The target is to get "15w)#
Testing
#2xx5 ->(5w+1)(2w+3) = 10w^2+15w+2w color(red)(larr" this fails")#
Testing
#2xx5 ->(2w+1)(5w+3) = 10w^2+6w+5w color(red)(larr" this fails")#
Testing
#1xx10 ->(w+1)(10w+3) = 10w^2+3w+10w color(red)(larr" this fails")#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(red)("This is not going to work so lets switch to the formula")#
#color(green)("Really worth memorising this one if you can.")#
Standard form using #y" and "x#
#y=ax^2+bx+c -> x=(-b+-sqrt(b^2-4ac))/(2a)#
Where: #a=10, b=15, c=3#
#y=(-15+-sqrt(15^2-4(10)(3)))/(2(10))#
#y=(-15+-sqrt(105))/20#
#y=-3/4+-sqrt(105)/20#
#y=-1.26" and "-.28# to 2 decimal places
