If the normal to the curve y(x)=∫_0^x(2 t^2-15 t +10) dt at a point ( a, b ) is parallel to the line x+3 y=-5, a >1, then the value of | a +6 b | is…

If the normal to the curve $\displaystyle y(x)=\int_0^x\left(2 \mathrm{t}^2-15 \mathrm{t}+10\right) \mathrm{dt}$ at a point $\displaystyle (\mathrm{a}, \mathrm{b})$ is parallel to the line $\displaystyle x+3 y=-5, \mathrm{a}>1$, then the value of $\displaystyle |\mathrm{a}+6 \mathrm{~b}|$ is equal to \_\_\_\_.