We have to solve the following expression:
[tex]\begin{gathered} \sqrt{x+83}=x+11 \\ x+83=(x+11)^2 \\ x+83=x^2+2*11*x+11^2 \\ x+83=x^2+22x+121 \\ 0=x^2+22x+121-x-83 \\ 0=x^2+21x+38 \end{gathered}[/tex]
We now have to apply the quadratic equation to find the possible solutions:
[tex]\begin{gathered} x=\frac{-21\pm\sqrt{21^2-4*1*38}}{2*1} \\ x=\frac{-21\pm\sqrt{441-152}}{2} \\ x=\frac{-21\pm\sqrt{289}}{2} \\ x=\frac{-21\pm17}{2} \\ =>x_1=\frac{-21-17}{2}=-\frac{38}{2}=-19 \\ =>x_2=\frac{-21+17}{2}=-\frac{4}{2}=-2 \end{gathered}[/tex]
We have found two possible solutions: x = -19 and x = -2.
One of this solution is valid and the other is not.
We can check this knowing that the square root will always (unless expressed the contrary with a minus sign) be a positive number.
This means that is √(x+83) is positive, so has to be x + 11.
The first solution is x = -19, so:
[tex]x+11=-19+11=-8<0[/tex]
Then, this solution is not the correct one.
We test the other solution (x = -2) and get:
[tex]x+11=-2+11=9>0[/tex]
This solution is the correct one.
Answer: x = -2.
