I have to solve this irrational inequality:

$\displaystyle \sqrt {25 - x^2 } + \sqrt {x^2 + 7x} > 3 $

Is this solution?

$\displaystyle \sqrt {25 - x^2 } > 0 \wedge \sqrt {x^2 + 7x} > 3 $

$\displaystyle \sqrt {25 - x^2 } > 1 \wedge \sqrt {x^2 + 7x} > 2 $

$\displaystyle \sqrt {25 - x^2 } > 2 \wedge \sqrt {x^2 + 7x} > 1 $

$\displaystyle \sqrt {25 - x^2 } > 3 \wedge \sqrt {x^2 + 7x} > 0 $