Calculate $\sqrt{x-1}$ to the power of $2$ and get $x-1$.
$$x-1=\left(x-7\right)^{2}$$
Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-7\right)^{2}$.
$$x-1=x^{2}-14x+49$$
Subtract $x^{2}$ from both sides.
$$x-1-x^{2}=-14x+49$$
Add $14x$ to both sides.
$$x-1-x^{2}+14x=49$$
Combine $x$ and $14x$ to get $15x$.
$$15x-1-x^{2}=49$$
Subtract $49$ from both sides.
$$15x-1-x^{2}-49=0$$
Subtract $49$ from $-1$ to get $-50$.
$$15x-50-x^{2}=0$$
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
$$-x^{2}+15x-50=0$$
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $-x^{2}+ax+bx-50$. To find $a$ and $b$, set up a system to be solved.
$$a+b=15$$ $$ab=-\left(-50\right)=50$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. List all such integer pairs that give product $50$.
$$1,50$$ $$2,25$$ $$5,10$$
Calculate the sum for each pair.
$$1+50=51$$ $$2+25=27$$ $$5+10=15$$
The solution is the pair that gives sum $15$.
$$a=10$$ $$b=5$$
Rewrite $-x^{2}+15x-50$ as $\left(-x^{2}+10x\right)+\left(5x-50\right)$.
$$\left(-x^{2}+10x\right)+\left(5x-50\right)$$
Factor out $-x$ in the first and $5$ in the second group.
$$-x\left(x-10\right)+5\left(x-10\right)$$
Factor out common term $x-10$ by using distributive property.
$$\left(x-10\right)\left(-x+5\right)$$
To find equation solutions, solve $x-10=0$ and $-x+5=0$.
$$x=10$$ $$x=5$$
Substitute $10$ for $x$ in the equation $\sqrt{x-1}=x-7$.
$$\sqrt{10-1}=10-7$$
Simplify. The value $x=10$ satisfies the equation.
$$3=3$$
Substitute $5$ for $x$ in the equation $\sqrt{x-1}=x-7$.
$$\sqrt{5-1}=5-7$$
Simplify. The value $x=5$ does not satisfy the equation because the left and the right hand side have opposite signs.
