So, $$\frac{\sqrt[2]{6x+6h+17}-\sqrt[2]{6x+17}}{h}$$
then this is where I am confused, not when multiplying by the conjugate to rationalize, the subtraction becomes addition.
$$\frac{\sqrt[2]{6x+6h+17}-\sqrt[2]{6x+17}}{h} \cdot \frac{\sqrt[2]{6x+6h+17}+\sqrt[2]{6x+17}}{\sqrt[2]{6x+6h+17}+\sqrt[2]{6x+17}}$$
then you can simply treat the numerator as raised to the power of two $$\frac{\left(\sqrt[2]{6x+6h+17}\right)^2-\left(\sqrt[2]{6x+17}\right)^2}{h\cdot\left(\sqrt[2]{6x+6h+17}+\sqrt[2]{6x+17}\right)}$$
simplify and cancel h
$$\frac{6h}{h\cdot\left(\sqrt[2]{6x+6h+17}+\sqrt[2]{6x+17}\right)}$$ $$\frac{6}{\sqrt[2]{6x+6h+17}+\sqrt[2]{6x+17}}$$
Why is the conjugate positive?