The isolation of x with squareroots

Let's start with the solving the general case, because we can then use that to solve the general problem.

$\sqrt{x+n} + \sqrt{x+m} = y$

Now let's start by squaring both sides. We will get.

$(x+n) + (x+m) + 2\sqrt{x+n}\sqrt{x+m} = y^{2}$

A bit of simplification gives us.

$2x + n + m + 2\sqrt{x+n}\sqrt{x+m} = y^{2}$

Here comes a hard step and it might sound weird. But we will now add a n to both sides and subtract m. Thus we get

$2x + n + m + 2\sqrt{x+n}\sqrt{x+m} + n - m= y^{2} + n - m$

And we will simplify again to get:

$2x + 2n + 2\sqrt{x+n}\sqrt{x+m} = y^{2} + n - m$

Now we will divide by 2.

$\frac{2x + 2n + 2\sqrt{x+n}\sqrt{x+m}}{2} = \frac{y^{2} + n - m}{2}$

And simplify to get

$x + n + \sqrt{x+n}\sqrt{x+m} = \frac{y^{2} + n - m}{2}$

Another bit tricky step, but we will pull out $\sqrt{x+n}$ out from parenthesis on the left side. This means, that we will get this:

$\sqrt{x+n}(\sqrt{x+n}+\sqrt{x+m}) = \frac{y^{2} + n - m}{2}$

However remember that: $\sqrt{x+n} + \sqrt{x+m} = y$ . That means we can replace it $\sqrt{x+n} + \sqrt{x+m}$ with $y$

$\sqrt{x+n}*(y) = \frac{y^{2} + n - m}{2}$

Now we can divide by y on both sides to get:

$\sqrt{x+n} = \frac{y^{2} + n - m}{2y}$

We can now square both sides.

$x+n = (\frac{y^{2} + n - m}{2y})^{2}$

And now to finish and isolate x, we will subtract n from both sides and get:

$x = (\frac{y^{2} + n - m}{2y})^{2} - n$

Now we can just replace the numbers in: $\sqrt{x+2} + \sqrt{x+3} = 5$ .

So we get that x is:

$x = (\frac{5^2+2-3}{2*5})^{2} - 2$

$x = 2.4^{2} - 2$

$x = 3.76$

Let $\sqrt{x+2}=a,\sqrt{x+3}=b$ ,then $a+b=5,b^2-a^2=1=(b+a)(b-a)=5(b-a),b-a=0.2,b=2.6,x=3.76$

Genaral case:

Let $\sqrt{x+n}=a,\sqrt{x+m}=b$ ,then $a+b=y,a^2-b^2=n-m=(a+b)(a-b)=y(a-b),a-b=\frac{n-m}y,a=\frac{y+\frac{n-m}y}2,x=a^2-n=(\frac{y^2+n-m}y)^2-n$

Generalization $\sqrt{x+m} + \sqrt {x+n} = y \implies \sqrt{x+m} = y-\sqrt {x+n}$ Squaring we have $x +m = y^2 -2y\sqrt{x+n} +x+n \implies \\ \sqrt{x+n} = \dfrac{y^2 +n-m}{2y} \implies x =\left(\dfrac{y^2+n-m}{2y}\right)^2-n \\ \therefore x = \left(\dfrac{5^2+3-2}{2\cdot 5}\right)^2-3 = \dfrac{169-75}{75} = \dfrac{94}{25}\approx \boxed{3.76 }$

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$\begin{aligned} \sqrt{x+n} + \sqrt{x+m} & = y & \small \color{#3D99F6} \text{Squaring both sides} \\ x + n + 2\sqrt{(x+n)(x+m)} + x + m & = y^2 & \small \color{#3D99F6} \text{Rearranging} \\ 2x + m + n - y^2 & = - 2\sqrt{(x+n)(x+m)} & \small \color{#3D99F6} \text{Squaring both sides} \\ 4x^2 + 4(m+n-y^2)x + (m+n-y^2)^2 & = 4x^2 + 4(m+n)x + 4mn & \small \color{#3D99F6} \text{Rearranging} \\ 4y^2x & = m^2+2mn + n^2 - 2(m+n)y^2 + y^4 - 4mn \\ x & = \frac {(m-n)^2 - 2(m+n)y^2+y^4}{4y^2} & \small \color{#3D99F6} \text{For }n =2, m=3, y = 5 \\ & = \frac {(2-3)^2-2(2+3)5^2+5^4}{4(5^2)} \\ & = \frac {376}{100} = \boxed{3.76} \end{aligned}$