Equation $\sqrt{5x^{2}-6x+8}-\sqrt{5x^{2}-6x-7}=1$ has
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All roots positive
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All roots negative
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One root positive and the other negative
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Two root positive and one root negative
Let y = 5x^2 - 6x. The equation is sqrt(y+8) - sqrt(y-7) = 1. Squaring gives (y+8) + (y-7) - 2*sqrt((y+8)(y-7)) = 1, so 2y + 1 - 1 = 2*sqrt((y+8)(y-7)). y = sqrt(y^2 + y - 56), y^2 = y^2 + y - 56, y = 56. 5x^2 - 6x - 56 = 0. Roots are (6 +/- sqrt(36 - 4*5*(-56))) / 10 = (6 +/- sqrt(1156)) / 10 = (6 +/- 34) / 10. Roots are 4 and -2.8. One positive, one negative.
Let u = sqrt(5x^2 - 6x + 8) and v = sqrt(5x^2 - 6x - 7), so the equation is u - v = 1 and squaring the definitions gives u^2 - v^2 = 15. Factoring the difference of squares yields (u - v)(u + v) = 15, so substituting u - v = 1 gives u + v = 15, which means u = 8 and v = 7. Substituting u back gives sqrt(5x^2 - 6x + 8) = 8, so 5x^2 - 6x + 8 = 64 simplifies to 5x^2 - 6x - 56 = 0. Using the quadratic formula, x = (6 +/- sqrt(36 + 1120))/10, yielding roots of (6 + 34)/10 = 4 and (6 - 34)/10 = -2.8, proving one root is positive and the other is negative.