Start with
[tex]3(x^2-13x+41)+8=11[/tex]
Subtract 8 from both sides:
[tex]3(x^2-13x+41)=3[/tex]
Divide both sides by 3:
[tex]x^2-13x+41=1[/tex]
Subtract 1 from both sides:
[tex]x^2-13x+40=0[/tex]
Solve with the usual formula
[tex]ax^2+bx+c=0\iff x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
to get
[tex]x^2-13x+40=0\iff x = \dfrac{13\pm\sqrt{9}}{2} = \dfrac{13\pm 3}{2}[/tex]
So, the two solutions are
[tex]x_1 = \dfrac{13+3}{2}=8,\quad x_2 = \dfrac{13-3}{2}=5[/tex]
