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\left(\sqrt{2x+22}\right)^{2}=\left(x+7\right)^{2}
Square both sides of the equation.
2x+22=\left(x+7\right)^{2}
Calculate \sqrt{2x+22} to the power of 2 and get 2x+22.
2x+22=x^{2}+14x+49
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+7\right)^{2}.
2x+22-x^{2}=14x+49
Subtract x^{2} from both sides.
2x+22-x^{2}-14x=49
Subtract 14x from both sides.
-12x+22-x^{2}=49
Combine 2x and -14x to get -12x.
-12x+22-x^{2}-49=0
Subtract 49 from both sides.
-12x-27-x^{2}=0
Subtract 49 from 22 to get -27.
-x^{2}-12x-27=0
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
a+b=-12 ab=-\left(-27\right)=27
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-27. To find a and b, set up a system to be solved.
-1,-27 -3,-9
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 27.
-1-27=-28 -3-9=-12
Calculate the sum for each pair.
a=-3 b=-9
The solution is the pair that gives sum -12.
\left(-x^{2}-3x\right)+\left(-9x-27\right)
Rewrite -x^{2}-12x-27 as \left(-x^{2}-3x\right)+\left(-9x-27\right).
x\left(-x-3\right)+9\left(-x-3\right)
Factor out x in the first and 9 in the second group.
\left(-x-3\right)\left(x+9\right)
Factor out common term -x-3 by using distributive property.
x=-3 x=-9
To find equation solutions, solve -x-3=0 and x+9=0.
\sqrt{2\left(-3\right)+22}=-3+7
Substitute -3 for x in the equation \sqrt{2x+22}=x+7.
4=4
Simplify. The value x=-3 satisfies the equation.
\sqrt{2\left(-9\right)+22}=-9+7
Substitute -9 for x in the equation \sqrt{2x+22}=x+7.
2=-2
Simplify. The value x=-9 does not satisfy the equation because the left and the right hand side have opposite signs.
x=-3
Equation \sqrt{2x+22}=x+7 has a unique solution.