\sqrt{x+5}+\sqrt{20-x} =7

Square both sides of the equation,

x+5+2\sqrt{x+5}\sqrt{20-x}+20-x=49

\sqrt{x+5}\sqrt{20-x}=12

Square both sides of the equation again,

(x+5)(20-x) = 144

x^2-15x+44=0

Factoring the quadratic expression on left hand side

(x-4)(x-11)=0

Now we have found the solution for the equation.

x=4 or x=11

\sqrt{x+5}+\sqrt{20-x} =7

Let

Then

Addition a^2 and b^2

a^2+b^2 = x+5+20-x=25

then

Squaring (3) gives

Subtract (4) from (5)

(4)+(6) gives

(a-b)^2 = 1

then

(3) and (7) gives two system of equaitons

\begin{cases} a+b&=7 \\ a-b &= 1 \end{cases}

and

\begin{cases} a+b&=7 \\ a-b &= -1 \end{cases}

Solving the equations yields

\begin{cases} a&=4 \\ b &= 3 \end{cases}

and

\begin{cases} a&=3 \\ b &= 4 \end{cases}

Substitute back to (1) and (2) and we get the solution to the equation

x = 4 or x = 11