How many real solutions of the equation `6x^(2)-77[x]+147=0`, where `[x]` is the integral part of `x`?

Answer

Four

Explanation

We have `6x^(2)-77[x]+147=0`
`implies(6x^(2)+147)/77=[x]`
`(0.078)x^(2)=[x]-1.9`
`:'(0.078)x^(2)gt0=x^(2)gt0`
`:.[x]-1.9gt0`
or `[x]gt1.9`
`:.[x]=2,3,4,5`…………
If `[x]=2` i.e. `2lexlt3`
Then `x^(2)=(2-1.9)/0.078=1.28`
`:.x=1.13` [fail]
If `[x]=3,` i.e. `3lexlt4`
Then `x^(2)=(3-1.9)/0.078=14.1`
`:.x=3.75` [true]
If `[x]=4`, i.e. `4lexlt5`
Then `x^(2)=(4-1.9)/(0.078)=26.9`
`:.x=5.18` ..[fail]
If `[x]=5,` i.e. `5lexlt6`
Then `x^(2)=(5-1.9)/0.078=39.7`
`:.x=6.3` [fail]
If `[x]=6,` i.e. `6lexlt7`
Then `x^(2)=(6-=1+N3269)/0.078=4.1/0.078=52.56`
`:.x==.25`[fail]
If `[x]=7`, i.e. /7lexlt8`
Then `x^(2)=(7-1.9)/0.078=5.1/0.078=65.38`
`:.x=8.08` [fail]
If `[x]=8` i.e. `8lexlt9`
Then `x^(2)=(8-1.9)/0.078=6.1/0.078=78.2`
`:.x=8.8` [true]
If `[x]=9,` i.e. `9lexlt10`
Then `x^(2)=(9-1.9)/0.078=7.1/0.078=91.03`
`:.x=9.5` [true]
If `[x]=10`, i.e `10lexlt11`
Then `x^(2)=(10-1.9)/0.078=8.1/0.078=103.8`
`:.x=10.2`
If `[x]=11`, i.e. `11lexlt12`
Then `x^(2)=(11-1.9)/(0.078`
`=9.1/0.078=116.7`
`:.x=10.8` [fail]
Other values of fail.
hence number of solutions is four.