Let \(t=\sqrt{x}\). Then \(t\ge 0\) and \(x=t^2\).

Substitute into the equation:

\(t^2-t=6\).

Move everything to one side:

\(t^2-t-6=0\).

Factorize:

\((t-3)(t+2)=0\).

Hence \(t=3\) or \(t=-2\).

But \(t=\sqrt{x}\ge 0\), so \(t=-2\) is impossible.

Therefore, \(t=3\), and \(x=3^2=9\).

Answer: \(9\).

Both sides of the equation are non-negative, so we can square both sides:

\(x^2-10=3x\).

Move everything to one side:

\(x^2-3x-10=0\).

Factorize:

\((x-5)(x+2)=0\).

We get the candidates \(x=5\) and \(x=-2\).

Check the domain: the radicand \(3x\ge 0\), so \(x\ge 0\).

Therefore, \(x=-2\) is impossible, and only \(x=5\) remains.

Answer: \(5\).

The right-hand side is \(x\), while the left-hand side is the principal square root, so \(x\ge 0\).

Square both sides:

\(2x+8=x^2\).

Move everything to one side:

\(x^2-2x-8=0\).

Factorize:

\((x-4)(x+2)=0\).

We get \(x=4\) or \(x=-2\).

Since \(x\ge 0\), the value \(-2\) is impossible.

Check: \(\sqrt{2\cdot4+8}=\sqrt{16}=4\).

Answer: \(4\).

A product is equal to zero if at least one of its factors is zero.

We also need the square root to exist: \(2x-2\ge 0\Rightarrow x\ge 1\).

1) \(\sqrt{2x-2}=0\Rightarrow 2x-2=0\Rightarrow x=1\).

2) \(x^2-25=0\Rightarrow x=\pm5\).

Taking into account \(x\ge 1\), only \(x=5\) is valid.

So the roots are \(1\) and \(5\).

According to the instruction, we write their sum:

\(1+5=6\).

Answer: \(6\).

First find the domain:

\(2x-1\ge 0\Rightarrow x\ge \dfrac12\).

Since both sides are non-negative, square the inequality:

\(2x-1\le 9\).

Hence:

\(2x\le 10\Rightarrow x\le 5\).

So, \(\dfrac12\le x\le 5\).

The integer values are \(1,2,3,4,5\).

Their number is \(5\).

Answer: \(5\).

First find the domain:

\(3x-2\ge 0\Rightarrow x\ge \dfrac23\),

\(x+12\ge 0\Rightarrow x\ge -12\).

So, \(x\ge \dfrac23\).

Since the square root function is increasing, compare the radicands:

\(3x-2\le x+12\).

\(2x\le 14\Rightarrow x\le 7\).

Thus, \(\dfrac23\le x\le 7\).

The integer solutions are \(1,2,3,4,5,6,7\).

Their number is \(7\).

Answer: \(7\).

Let \(t=\sqrt{x}\). Then \(t\ge 0\) and \(x=t^2\).

Substitute:

\(t^2-2t=15\).

Move everything to one side:

\(t^2-2t-15=0\).

Factorize:

\((t-5)(t+3)=0\).

We get \(t=5\) or \(t=-3\).

Since \(t\ge 0\), only \(t=5\) remains.

Then \(x=25\).

Answer: \(25\).

Both sides are non-negative, so square both sides:

\(x^2-3=2x\).

We get the quadratic equation:

\(x^2-2x-3=0\).

Factorize:

\((x-3)(x+1)=0\).

The candidates are \(x=3\) and \(x=-1\).

But from \(2x\ge 0\), we have \(x\ge 0\).

Therefore, \(x=-1\) is impossible, and \(x=3\) remains.

Answer: \(3\).

Since \(\sqrt{4x-3}=x\), we must have \(x\ge 0\).

Square both sides:

\(4x-3=x^2\).

Then:

\(x^2-4x+3=0\).

Factorize:

\((x-1)(x-3)=0\).

Hence \(x=1\) or \(x=3\).

Both values satisfy the original equation.

According to the instruction, we write the sum of the roots:

\(1+3=4\).

Answer: \(4\).

A product is zero if at least one factor is zero.

We also need the square root to exist: \(3x+3\ge 0\Rightarrow x\ge -1\).

1) \(\sqrt{3x+3}=0\Rightarrow 3x+3=0\Rightarrow x=-1\).

2) \(x^2-36=0\Rightarrow x=\pm6\).

Taking into account \(x\ge -1\), only \(x=6\) is valid; \(-6\) is impossible.

So the roots are \(-1\) and \(6\).

According to the instruction, write their sum:

\(-1+6=5\).

Answer: \(5\).

First find the domain:

\(2x-5\ge 0\Rightarrow x\ge \dfrac52\).

Since both sides are non-negative, square the inequality:

\(2x-5\le 4\).

Hence:

\(2x\le 9\Rightarrow x\le \dfrac92\).

So, \(\dfrac52\le x\le \dfrac92\).

The integer solutions are \(3\) and \(4\).

Their number is \(2\).

Answer: \(2\).

Find the domain:

\(2x+1\ge 0\Rightarrow x\ge -\dfrac12\),

\(x+8\ge 0\Rightarrow x\ge -8\).

Therefore, \(x\ge -\dfrac12\).

Since the square root function is increasing, compare the radicands:

\(2x+1\le x+8\).

Hence:

\(x\le 7\).

So, \(-\dfrac12\le x\le 7\).

The integer solutions are \(0,1,2,3,4,5,6,7\).

Their number is \(8\).

Answer: \(8\).

Let \(t=\sqrt{x}\). Then \(t\ge 0\) and \(x=t^2\).

We get the equation:

\(t^2-5t=14\).

Move everything to one side:

\(t^2-5t-14=0\).

Factorize:

\((t-7)(t+2)=0\).

Hence \(t=7\) or \(t=-2\).

Since \(t\ge 0\), only \(t=7\) remains.

Then \(x=49\).

Answer: \(49\).

Square both sides:

\(x^2-7=6x\).

Move everything to one side:

\(x^2-6x-7=0\).

Factorize:

\((x-7)(x+1)=0\).

The candidates are \(x=7\) and \(x=-1\).

But from \(6x\ge 0\), that is, \(x\ge 0\), the value \(-1\) is impossible.

Therefore, only \(x=7\) remains.

Answer: \(7\).

Since \(\sqrt{4x+12}=x\), we must have \(x\ge 0\).

Square both sides:

\(4x+12=x^2\).

Move everything to one side:

\(x^2-4x-12=0\).

Factorize:

\((x-6)(x+2)=0\).

Hence \(x=6\) or \(x=-2\).

Since \(x\ge 0\), the value \(-2\) is impossible.

Check: \(\sqrt{4\cdot6+12}=\sqrt{36}=6\).

Answer: \(6\).

A product is zero if at least one factor is zero.

Domain: \(5x-10\ge 0\Rightarrow x\ge 2\).

1) \(\sqrt{5x-10}=0\Rightarrow 5x-10=0\Rightarrow x=2\).

2) \(x^2-64=0\Rightarrow x=\pm8\).

Taking into account \(x\ge 2\), only \(x=8\) is valid; \(-8\) is impossible.

The roots are \(2\) and \(8\).

According to the instruction, write their sum:

\(2+8=10\).

Answer: \(10\).