Question 990409
Using the first two sides,
{{{x+12=3x+8}}}
{{{2x=4}}}
{{{x=2}}}
The second and the last side,
{{{3x+8=2x+2}}}
{{{x=-6}}}
.
.
This triangle can't be equilateral.
The sides must be the same so x would always come out the same independent of which pair of sides you choose.
Please check the problem set up and repost.