**The Shorter and Easier Way**

We see that there are 2 #(-2y-3)# terms in the equation. This might signal something useful. So, let's temporarily replace #(-2y-3)# with #x#. Now, let's rewrite the equation:

#x^3+25x=0#

Using the Distributive Property, we can factor out an #x# from the left side:

#x*(x^2+25)=0#

Looking at our new equation, we can see that the two terms' products is #0#. This means that either #x# is #0# or #x^2+25# is #0#. Let's start with the case of #x=0# and solve for #y#. However, before we start, remember that #x# is actually #(-2y-3)#:

#x=0#

#-2y-3=0#

#-2y=3#

#y=-3/2#

Besides this single answer of #y=-3/2#, there are two more roots (solutions) which involve complex numbers. To obtain these two other solutions, we have to look at the case where #x^2+25=0#.

#x^2+25=0#

#x^2=-25#

#x=sqrt-25#

#x=sqrt-1*sqrt25#

#x=+-5i#

The final statement shows that #x# can be either positive #5i# of negative #5i#. We'll start by working with positive #5i#. And, because we let #x=(-2y-3)#, we can substitute it in and find the second value for #y#:

#-2y-3=5i#

#-2y=5i+3#

#y=(5i+3)/-2#

#y=(5i)/(-2)+3/(-2)#

#y=(-5/2)i-3/2#

#y=-3/2-(5/2)i#

The third solution for #y# involves seeing #x# as equal to #-5i#:

#-2y-3=-5i#

#-2y=-5i+3#

#y=(-5i+3)/-2#

#y=(-5i)/-2+3/-2#

#y=-3/2+(5/2)i#

**The Long and Hard Way**

Let's start with the original problem:

#(-2y-3)^3+25(-2y-3)=0#

Let's expand everything out and use the distributive property to simplify things a bit:

#(-2y-3)(-2y-3)(-2y-3)+(25)(-2y)-(25)(3)=0#

#(-2y-3)(4y^2+6y+6y+9)+(-50y)-75=0#

#(4y^2)(-2y-3)+(6y)(-2y-3)+(6y)(-2y-3)+(9)(-2y-3)-50y-75=0#

#(-8y^3-12y^2)+(-12y^2-18y)+(-12y^2-18y)+(-18y-27)-50y-75=0#

Now let's break open the parentheses:

#-8y^3-12y^2-12y^2-18y-12y^2-18y-18y-27-50y-75=0#

And now, let's start rearranging like terms together and simplifying them:

#-8y^3-12y^2-12y^2-12y^2-18y-18y-18y-50y-27-75=0#

#-8y^3-36y^2-104y-102=0#

At this point, we can use a cubic calculator to calculate the value of #y#:

#y=-1.5#

**It's important that two other solutions also work perfectly fine, not just #y=-1.5#. The two other solutions are: #y=-3/2+5/2i# and #y=-3/2-5/2i#.**

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Checking:

If we were to check whether or not #y# is indeed #-1.5#, we would substitute #y=-1.5# back into the original equation:

#[-2(-1.5)-3]^3+25[-2(-1.5)-3]=0#

#(3-3)^3+25(3-3)=0#

#0^3+25(0)=0#

#0+0=0#

#0=0#