Х+√(2х²-14х+13)=5 помогиие пешить

$x+ \sqrt{2x^2-14x+13}=5 \\ \sqrt{2x^2-14x+13}=5-x \\ (\sqrt{2x^2-14x+13})^2=(5-x)^2 \\ 2x^2-14x+13=25-10x+ x^{2} \\ 2 x^{2} - x^{2} -14x+10x+13-25=0 \\ x^{2} -4x-12x=0 \\ \\ D=(-4)^2-4*(-12)*1=16+48=64=8^2 \\ x_1= \frac{-(-4)- \sqrt{64} }{2*1}= \frac{4-8}{2} =- \frac{4}{2}=-2 \\ x_2= \frac{-(-4)+\sqrt{64} }{2*1}= \frac{4+8}{2} =\frac{12}{2}=6 \\ \\ -2+ \sqrt{2*(-2)^2-14(-2)+13}=-2+ \sqrt{2*4+28+13} = \\ =-2+ \sqrt{8+41} =-2+ \sqrt{49} =-2+7=5 \\$
$6+ \sqrt{2*6^2-14*6+13}=6+ \sqrt{2*36-84+13}=6+ \sqrt{72-71}= \\ =6+ \sqrt{1} =7 \\ 7 \neq 5$
Ответ: х= -2