Question

The root of the equation 2y²-3y-2=0 are ----

Answer 1

2y^2 - 3y - 2= 0

2y^2 -4y + y - 2 = 0

2y(y-2) +1(y-2) = 0

(2y+1)(y-2) = 0

=> (2y+1) = 0. or. (y-2) = 0

=> y = -1/2. or. y = 2

Hence, roots of the equation are -1/2 and 2.