Question 510706
The problem, as stated, is not solvable!
If the triangle is a right triangle, or an equilateral triangle, you have to say so. Let's assume that it's a right triangle so we can use the Pythagorean theorem: {{{c^2 = a^2+b^2}}}
For any positive value of x, 2x+1 has to be the largest side and thus the hypotenuse (c) so...
{{{(2x+1)^2 = (x-5)^2+(2x)^2}}} Simplify.
{{{4x^2+4x+1 = x^2-10x+25+4x^2}}}
{{{4x^2+4x+1 = 5x^2-10x+25}}} Subtract 4x^2 from both sides.
{{{4x+1 = x^2-10x+25}}} Subtract 4x from both sides.
{{{1 = x^2-14x+25}}} Subtract 1 from both sides.
{{{x^2-14x+24 = 0}}} Solve by factoring.
{{{(x-2)(x-12) = 0}}} Apply the zero product rule.
{{{x-2 = 0}}} or {{{x-2 = 0}}}
{{{x = 2}}} or {{{x = 12}}} Discard {{{x = 2}}} as an invalid solution because x = 2 will give you a negative (x-5 = -3) side length.
Answer is {{{x = 12}}}