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\sqrt{2x+26}=x+9
Subtract -9 from both sides of the equation.
\left(\sqrt{2x+26}\right)^{2}=\left(x+9\right)^{2}
Square both sides of the equation.
2x+26=\left(x+9\right)^{2}
Calculate \sqrt{2x+26} to the power of 2 and get 2x+26.
2x+26=x^{2}+18x+81
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+9\right)^{2}.
2x+26-x^{2}=18x+81
Subtract x^{2} from both sides.
2x+26-x^{2}-18x=81
Subtract 18x from both sides.
-16x+26-x^{2}=81
Combine 2x and -18x to get -16x.
-16x+26-x^{2}-81=0
Subtract 81 from both sides.
-16x-55-x^{2}=0
Subtract 81 from 26 to get -55.
-x^{2}-16x-55=0
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
a+b=-16 ab=-\left(-55\right)=55
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-55. To find a and b, set up a system to be solved.
-1,-55 -5,-11
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 55.
-1-55=-56 -5-11=-16
Calculate the sum for each pair.
a=-5 b=-11
The solution is the pair that gives sum -16.
\left(-x^{2}-5x\right)+\left(-11x-55\right)
Rewrite -x^{2}-16x-55 as \left(-x^{2}-5x\right)+\left(-11x-55\right).
x\left(-x-5\right)+11\left(-x-5\right)
Factor out x in the first and 11 in the second group.
\left(-x-5\right)\left(x+11\right)
Factor out common term -x-5 by using distributive property.
x=-5 x=-11
To find equation solutions, solve -x-5=0 and x+11=0.
\sqrt{2\left(-5\right)+26}-9=-5
Substitute -5 for x in the equation \sqrt{2x+26}-9=x.
-5=-5
Simplify. The value x=-5 satisfies the equation.
\sqrt{2\left(-11\right)+26}-9=-11
Substitute -11 for x in the equation \sqrt{2x+26}-9=x.
-7=-11
Simplify. The value x=-11 does not satisfy the equation.
x=-5
Equation \sqrt{2x+26}=x+9 has a unique solution.