Find the turning points of the curve y = x^3 +5x^2 -6x +4

y= x³ +5x² -6x +4

dy/dx = 3x² +10x -6

at turning points dy/dx = 0 therefore 

3x² +10x -6 = 0

This quadratic is factorisable. When factorised you get:

(3x -2)(x +4) = 0

therefore x = 2/3 and -4 at the turning points

to find the y co-ordinates, substitue these values of x into the original equation of y= x^3 +5x^2 -6x +4

y = (-4)³ +5(-4)² -6(-4) +4 = 44

y = (2/3)³ +5(2/3)² -6(2/3) +4 = 68/27

thw turning points of the curve are at the points (-4,44) and (2/3,68/27)