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]