Solve the equation `2x^(2)+x-4=0` by the method of completing the square.

Explanation

We have
`2x^(2)+x-4=0`
`implies" "4x^(2)+2x-8=0" "["multiplying both sides by "2]`
`implies" "4x^(2)+2x=8`
`implies" "(2x)^(2)+2xx2x xx(1)/(2)+((1)/(2))^(2)=8+((1)/(2))^(2)`
`" ""[""adding "((1)/(2))^(2)" on both sides""]"`
`implies" "(2x+(1)/(2))^(2)=(8+(1)/(4))=(33)/(4)=((sqrt(33))/(2))^(2)`
`implies" "2x+(1)/(2)=+-((sqrt(33))/(2))" "["taking square root on both sides"]`
`implies" "2x+(1)/(2)=(sqrt(33))/(2)" or "2x+(1)/(2)=(-sqer(33))/(2)`
`implies" "2x=((sqrt(33))/(2)-(1)/(2))=((sqrt(33)-1))/(2)`
or`" "2x=(-sqrt(33))/(2)-(1)/(2)=(-"("sqrt(33)+1")")/(2)`
`implies" "x=("("sqrt(33)-1")")/(4)" or "x=(-"("sqrt(33)+1")")/(4).`
Hence, `(-1+sqrt(33))/(4)" and "(-1-sqrt(33))/(4)` are the roots of the given equation.