Question 258088
<pre><font size = 4 color = "indigo"><b>
Heron's formula for the area A of a triangle with
sides a,b, and c is

{{{A=sqrt(s(s-a)(s-b)(s-c))}}} where {{{s=(a+b+c)/2}}}

So we have a=2, b=5, and c=x and A = x.

We will eventually let {{{a=2}}}, {{{b=5}}}, {{{c=x}}},
and {{{A=x}}}

But to make things easier we will first simplify
Heron's formula by first removing the square root and
secondly, eliminating s:

{{{A=sqrt(s(s-a)(s-b)(s-c))}}}

{{{A^2=s(s-a)(s-b)(s-c)}}}

where

{{{s-a=(a+b+c)/2-a=(a+b+c)/2-2a/2=(a+b+c-2a)/2=(-a+b+c)/2}}}
{{{s-b=(a+b+c)/2-b=(a+b+c)/2-2b/2=(a+b+c-2b)/2=(a-b+c)/2}}}
{{{s-c=(a+b+c)/2-c=(a+b+c)/2-2c/2=(a+b+c-2c)/2=(a+b-c)/2}}}

so, substituting to eliminate s,

{{{A^2=s(s-a)(s-b)(s-c)}}}

{{{A^2=((a+b+c)/2)((-a+b+c)/2)((a-b+c)/2)((a+b-c)/2)}}}

{{{A^2=((a+b+c)(-a+b+c)(a-b+c)(a+b-c))/16}}}

{{{16A^2=(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}}

Now that we have rewritten Heron's formula in terms of the
sides only, we simplify the factors to substitute in
the above revised version of Heron's formula, which I think
is superior to the original version, especially for 
calculating purposes.

So now we substitute {{{a=2}}}, {{{b=5}}}, {{{c=x}}}, and {{{A=x}}}

{{{A^2=(5)^2=25}}}
{{{a+b+c = 2+5+x=7+x}}}
{{{-a+b+c = -2+5+x=3+x}}}
{{{a-b+c = 2-5+x=-3+x}}}
{{{a+b-c = 2+5-x=7-x}}}

Substituting these factors:

{{{16A^2=(a+b+c)(a+b-c)(a-b+c)(-a+b+c))}}}

{{{16(25)=(7+x)(3+x)(-3+x)(7-x)}}}

{{{400=(7+x)(7-x)(3+x)(-3+x)}}}

{{{400=(49-x^2)(-9+x^2)}}}

{{{400=-441+58x-x^4}}}

{{{841=58x-x^4)}}}

{{{x^4-58x+841=0}}}

That factors as

{{{(x-29)(x-29)=0}}}

So {{{x-29=0}}}

and {{{x=29}}}

Edwin</pre>