Find the equation of tangents to the curve y=x³+2x, which are perpendicular to the line x+14y+3=0.

The slope of line x+14y+3=0 is -1/14, so the slope of line perpendicular to given line is 14. Given curve is y=x³+2x, the slope of tangent to given curve is dy/dx=3x²+2=14. We get x²=4, which gives x=±2. At x=2, we get y=8 and at x=-2, we get y=-8. Therefore there are two tangents having slope 14 and one passing through (2,8) and the other passing through (-2,-8). The equation of tangent passing through point (2,8) is y-8=14(x-2), which gives 14x-y-16=0. The equation of line passing through point (-2,-8) is y+8=14(x+2), which gives 14x-y+16=0.