p (q(x)) = √ [ - ( 8x^2 + 10x - 3) ]
For the domain to be defined, - (8x^2 + 10x - 3 ) must be ≥ 0
So
-(8x^2 + 10x - 3) ≥ 0
-8x^2 - 10x + 3 ≥ 0
To solve this, set this = 0 and solve for x
-8x^2 - 10x + 3 = 0 multiply through by -1
8x^2 + 10x - 3 = 0 factor
(4x - 1) ( 2x + 3) = 0
Set each factor to 0 and solve for x
4x - 1 = 0 and 2x + 3 = 0
4x = 1 2x = - 3
x = 1/4 x = - 3/2
The solution will either come from x = (-inf, -3/2] U [1/4, infinity) or x = [-3/2, 1/4]
Testing a point in the second interval [ I'll pick x = 0 ].....we can see that
√ [ - ( 8(0)^2 + 10(0) - 3) ] = √ [ - ( -3) ] = √ 3 gives us a real number
So....the interval that solves this is [ -3/2, 1/4 ]
So b - a = 1/4 - (-3/2) = (1/4) - (-6/4) = 7/4
Here's a graph to show that the solution interval is correct :
https://www.desmos.com/calculator/6jhmgavfr0