Question 762677
use the determinant b^2-4ac
which is 
(-0.7)^2  - 4 x 0.25 x 1.5
this = -1.01
when the determinant is less then 0 then the roots are not real.

the points A and B obviously are on both the line and the curve.  to find them we neeed to put 2x -5 in the equation of the curve whenever it has y

x^2 + (2x-5)x -2 =0

i have then -2 on each side so i can equal the equation to 0.  the equation is now
x^2 + 2x^2 -5x -2=0
3x^2 -5x -2 =0

now find the roots of this equation
i used the formula to get x=2 and x= -1/3

put these values into any of the original equtions to get the y co-ordinates.

y=2x-5, so when x = 2 y = -1 and when x=-1/3 y=-17/3.  these are the points A and B