When factorising a quadratic that has a coefficient #>1#, we always multiply the coefficient by the constant given at the end of the quadratic.

#therefore# #2y^2+29y+90 -> 2y^2+29y+180#

When factorising, we always want to find two numbers that add to make the second number #(29)# and multiply to make #(180)#. The best way to do this is to find the factors of the constant at the end of the quadratic #(180)# and see if they can add to make the middle number #(29)#

As this quadratic has a #+# and another #+# this means both numbers will be positive.

#180:#

#180xx1# cannot make to #29#...

#90xx2# cannot make #29#...

#60xx3# cannot make #29#...

#45xx4# cannot make #29#...

#36xx5# cannot make #29#...

#30xx6# cannot make #29#...

#20xx9# CAN make #29#... so therefore we use these two numbers.

Removing the #29# and plugging the #20# and #9# into the quadratic:

#2x^2+20x+9x+90#

Always remember to change the constant back to the original...

Split up into two parts:

#2x^2+20x# and #9x+90#

#2x^2+20x -> 2x(x+10)# taking out a factor of #2x#

#9x+90 -> 9(x+10)# taking out a factor of #9#

This leaves us with:

#2x(x+10)+9(x+10)# <- Notice that the brackets are the same, this is one of our solutions. The brackets should ALWAYS be the same...

#2xcancel((x+10))+9cancel((x+10))#

This leaves us with our other bracket of #(2x+9)#

#therefore# the final answer is #(x+10)(2x+9)#

We can always check by expanding out the brackets:

#2x^2+9x+20x+90 -> 2x^2+29x+90#

#therefore# #(x+10)(2x+9)# is correct...