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\left(\sqrt{7x-34}\right)^{2}=\left(x-6\right)^{2}
Square both sides of the equation.
7x-34=\left(x-6\right)^{2}
Calculate \sqrt{7x-34} to the power of 2 and get 7x-34.
7x-34=x^{2}-12x+36
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-6\right)^{2}.
7x-34-x^{2}=-12x+36
Subtract x^{2} from both sides.
7x-34-x^{2}+12x=36
Add 12x to both sides.
19x-34-x^{2}=36
Combine 7x and 12x to get 19x.
19x-34-x^{2}-36=0
Subtract 36 from both sides.
19x-70-x^{2}=0
Subtract 36 from -34 to get -70.
-x^{2}+19x-70=0
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
a+b=19 ab=-\left(-70\right)=70
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-70. To find a and b, set up a system to be solved.
1,70 2,35 5,14 7,10
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 70.
1+70=71 2+35=37 5+14=19 7+10=17
Calculate the sum for each pair.
a=14 b=5
The solution is the pair that gives sum 19.
\left(-x^{2}+14x\right)+\left(5x-70\right)
Rewrite -x^{2}+19x-70 as \left(-x^{2}+14x\right)+\left(5x-70\right).
-x\left(x-14\right)+5\left(x-14\right)
Factor out -x in the first and 5 in the second group.
\left(x-14\right)\left(-x+5\right)
Factor out common term x-14 by using distributive property.
x=14 x=5
To find equation solutions, solve x-14=0 and -x+5=0.
\sqrt{7\times 14-34}=14-6
Substitute 14 for x in the equation \sqrt{7x-34}=x-6.
8=8
Simplify. The value x=14 satisfies the equation.
\sqrt{7\times 5-34}=5-6
Substitute 5 for x in the equation \sqrt{7x-34}=x-6.
1=-1
Simplify. The value x=5 does not satisfy the equation because the left and the right hand side have opposite signs.
x=14
Equation \sqrt{7x-34}=x-6 has a unique solution.