Is it possible to solve an algebraic equation in R?

You can use polynom package:

library(polynom)
p <- polynomial(c(44,51,6,-1))
# 44 + 51*x + 6*x^2 - x^3 
solve(p)
# [1] -4 -1 11

But you simply can use the function polyroot from base package:

polyroot(c(44,51,6,-1))
# [1] -1+0i -4+0i 11+0i

If you keep the real part with Re:

Re(polyroot(c(44,51,6,-1)))
# [1] -1 -4 11

Here we solve for the roots using the relationship between a matrix and its characteristic polynomial.

Given the polynomial a0 + a1*x^1 + a2*x^2 + x^3, define the matrix:

0   0  -a0
1   0  -a1
0   1  -a2

The eigenvalues of this matrix are the roots of the polynomial.

Substituting y = -x in your polynomial equation gives this

y^3 + 6*y^2 - 51*y + 44=0

And gives this example

> z <- matrix(c(0,1,0,0,0,1,-44,51,-6),3,3)
> z
     [,1] [,2] [,3]
[1,]    0    0  -44
[2,]    1    0   51
[3,]    0    1   -6
> eigen(z)
$values
[1] -11   4   1

$vectors
           [,1]        [,2]        [,3]
[1,]  0.6172134  0.73827166  0.98733164
[2,] -0.7715167 -0.67115606 -0.15707549
[3,]  0.1543033 -0.06711561 -0.02243936

Or, since we've substituted -y for x:

> eigen(-z)$values
[1] 11 -4 -1

See: http://www-math.mit.edu/~edelman/publications/polynomial_roots.pdf

I just stumbled upon this question and I am not sure if anything inherently changed around the Ryacas package, but it seems to work great in 2020, here is a helpful vignette to get started: https://cran.r-project.org/web/packages/Ryacas/vignettes/getting-started.html

Following the vignette, things seem to work as expected when I run the code:

library(Ryacas)
# initialize equation:
eq <- "-x^3+6*x^2+51*x+44"
# simplify the equation:
library(glue)
yac_str(glue("Simplify({eq})"))

[1] "6*x^2-x^3+51*x+44"

# factor:
yac_str(glue("Factor({eq})"))

[1] "(-1)*(x-11)*(x+4)*(x+1)"

You can evaluate the expression like this, plugging in whatever values for x:

# evaluate
evaluate(eq,list(x=c(0,1,10,100,-100)))

[[1]]
$src
[1] "-x^3+6*x^2+51*x+44"

attr(,"class")
[1] "source"

[[2]]
[1] "[1]      44     100     154 -934856 1054944\n"

Here you can see the results where x=0 produced an answer of 44, x=1 produced an answer of 100, etc...

If you evaluated the new simplified or factored versions and evaluated those, you would of course end up with the same exact results:

evaluate(yac_str(glue("Simplify({eq})")),list(x=c(0,1,10,100,-100)))

[[1]]
$src
[1] "6*x^2-x^3+51*x+44"

attr(,"class")
[1] "source"

[[2]]
[1] "[1]      44     100     154 -934856 1054944\n"

Notice the formula changed in the $src output, but we get the same results.

Here's the factored one too: evaluate(yac_str(glue("Factor({eq})")),list(x=c(0,1,10,100,-100)))

[[1]]
$src
[1] "(-1)*(x-11)*(x+4)*(x+1)"

attr(,"class")
[1] "source"

[[2]]
[1] "[1]      44     100     154 -934856 1054944\n"

The only real difference between what I outlined here and what's outlined in the vignette is the actual formula, and the fact that I used library(glue) instead of paste0, which is also a fair option.