Найдите x+12 , если кореньx+1 +x-11=0

[latex] \sqrt{x+1} +x-11=0 \\ \sqrt{x+1}=11-x \\ x+1=121-22x+ x^{2} \\ x^{2} -23x+120=0 \\ D=23^{2}-4*120=529-480=49=7^{2} \\ x_{1}= \frac{-(-23)-7}{2} = \frac{23-7}{2} = \frac{16}{2} =8 \\ x_{2}= \frac{-(-23)+7}{2} = \frac{23+7}{2} = \frac{30}{2} =15 \\ \\ \sqrt{8+1} +8-11= \sqrt{9} -3=3-3=0 \\ \sqrt{15+1} +15-11= \sqrt{16} +4=4+4=8 \\ x=8\\ 8+12=20[/latex]