Since the point P(a, b) lies on the curve: y=2x^2-3x+1 hence it will satisfy the equation of curve as follows
b=2a^2-3a+1\ ...........(1)
The slope m of the tangent line parallel to the given line: y=5x will be equal to the slope of given line i.e.
m=5
Now, the equation of the tangent line with slope m=5 & drawn at P(a, b)
y-b=5(x-a)
y=5(x-a)+b\ ........(2)
substituting y=5(x-a)+b in the equation of curve: y=2x^2-3x+1, we get
5(x-a)+b=2x^2-3x+1
2x^2-8x+5a-b+1=0
Since, the line: y=5(x-a)+b is tangent to the given curve at a single point hence above equation will have equal real roots i.e. determinant b^2-4ac of above quadratic equation will be 0 as follows
(-8)^2-4(2)(5a-b+1)=0
b=5a-7\ ........(3)
equating (1) & (3), we get
2a^2-3a+1=5a-7
a^2-4a+4=0
(a-2)^2=0
a=2, 2
substituting a=2 in (2), we get
b=5(2)-7=3
substituting values of a=2 & b=3 in (2), we get equation of tangent
y=5(x-2)+3
y=5x-7
Since, above tangent intersects x-axis at (d, 0) hence this point will satisfy above equation
0=5d-7
d=7/5