I think you meant to say the equation is
y = 2x² - 2x + 3
Differentiate both sides with respect to x :
dy/dx = 4x - 2
At the point (3, 0), the slope of the tangent line is dy/dx(3) = 4•3 - 2 = 10. Then the normal line to the curve at (3, 0) has slope -1/10.
Using the point-slope formula, the equation of the normal line is
y - 0 = -1/10 (x - 3) ⇒ y = (3 - x)/10