y'=6x^2-1, so y'=5 at (1,6). The slope of the tangent is therefore 5 and the equation of the tangent is y=mx+c where m=5. To find c, substitute (1,6) in this line 6=5+c, so c=1 and y=5x+1 is the equation of the tangent. The equation of the normal has slope -1/5 (-1/m) and equation y=-x/5+k, where k can be found by substituting (1,6) again: 6=-1/5+k, so k=31/5 and 5y=31-x is the equation of the normal (y=-x/5+31/5 is the same as 5y=-x+31 or 31-x, because we just multiply through by 5 to get rid of the fraction).