What is the equation of a line that passes through the point (1, 8) and is perpendicular to the line whose equation is y=x2+3 ?

Important:  If you're discussing squares, please express "the square of x" as x^2 (not as x2).  

Please note that your y=x^2+3 is NOT a line.  I will assume (probably incorrectly!) that you meant y=2x+3.  The slope of this line is 2, so the slope of any line perpendicular to y=2x+3 is -1/2.

Now use the point-slope equation of a straight line:

y-8 = (-1/2)(x-1)   (answer)