Solving for height “b” to satisfy back to back pythagoras triagles Hi I am have a real life problem, but somehow cannot get the 'b' side of the equation to work, and think I am making the wrong assumption for the "c's". I hope someone would be so kind to point out where my problem is and how to solve it.
I have attached the model of my problem, and I know the following:

-The known a's: Hin, Hout
The b's: (b+Vin), (b+Vout), where I know the Vin and Vout, and am trying to solve for b that satisfies both triangles. Where a V value is negative, I would put a negative value in the equation. 
The c's: I know the total length of the two summed hypotenuses, Ctot, but not the individual Cin and Cout lengths
And knowing the Pythagoras theorem, and rearranging to get a quadratic equation I hope I can find b. But, I think I am making the wrong assumption for the formula for the Ctot, or something else. I need help on this part of the assumption for my formula, as I cannot see where I am going wrong. I have redone this a few times, and get the same result each time.
This is what I got: $\sqrt{Hin^2+(b+Vin)^2}+\sqrt{Hout^2+(b+Vout)^2}=Ctot$
Where the first square root is my algebraic representation of the first, In, triangle. And, the second square root is the Out triangle. The two hypotenuses lengths sum up to my total length of Ctot
Combining two Pythagoras equations, and making them equal 0 to get a quadratic formula, I get: $(Hin^2+(b+Vin)^2)+(Hout^2+(b+Vout)^2)−Ctot^2=0$
Expanding I get the below, $(2∗b^2)+(2∗(Vin+Vout)∗b)+(Vin^2+Hin^2+Vout^2+Hout^2 −Ctot^2$) 
For my quadratic equation I get
$\frac{−2∗(Vin+Vout)+\sqrt{(2∗(Vin+Vout)^2−4(2)(Vin^2+Hin^2+Vout^2+Hout^2−Ctot^2)}}{2∗2}$  
When I use the Quadratic Formula to solve for b, and substitute into the two Pythagorean formulas to find the individual c's, of Cin and Cout, and sum them, they don't add up to my Ctot.
I need some help, from someone else's perspective, to see where I am going wrong. To me everything seems right, but the end result does not add up. 
Regards and many thank-you in advance,
Miguel
 A: 
This is what I got: $\;\sqrt{H_{in}^2+(b+V_{in})^2}+\sqrt{H_{out}^2+(b+V_{out})^2}=C_{tot}$

The second term should rather be $\,\sqrt{H_{out}^2+(b\color{red}{-}V_{out})^2}\,$ according to the posted diagram.

Combining two Pythagoras equations, and making them equal 0 to get a quadratic formula, I get: $(H_{in}^2+(b+V_{in})^2)+(H_{out}^2+(b+V_{out})^2)−C_{tot}^2=0$

This is wrong. You appear to be squaring $\,x+y=z\,$ term-by-term to "get" $\,x^2+y^2=z^2\,$, but it doesn't work that way. Instead, you should be getting $\,(x+y)^2=z^2 \iff x^2+2xy+y^2=z^2\,$:
$$
H_{in}^2+(b+V_{in})^2+H_{out}^2+(b-V_{out})^2 \color{red}{+ 2 \sqrt{(H_{in}^2+(b+V_{in})^2) \cdot(H_{out}^2+(b-V_{out})^2)}}= C_{tot}^2
$$
Isolating the $\color{red}{\style{font-family:inherit}{\text{red}}}$ term on one side and squaring again will yield a $\,4^{th}\,$ degree equation in $\,b\,$. While that's technically solvable by radicals, in case of a "real life problem" it would probably make more sense to solve it numerically.
