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Note that

$\sqrt{3x^2 + 6x + 7}$ = $\sqrt{3(x + 1)^2 + 4}$ ≥ $\sqrt{4}$ = $2$ ,

$\sqrt{5x^2 + 10x + 14}$ = $\sqrt{5(x+1)^2 + 9}$ ≥ $\sqrt{9}$ = $3$ ,

$\ 4 - 2x - x^2$ = $\ 5 - (x+1)^2$ ≤ 5.

consequently, the left hand side of the equation ≥ 5 for any $x$ belonging to R and the right hand side ≤ 5 for any $x$ belonging to R. This equality is only possible under the condition that both sides of the equation are equal to $5$ for the same value of $x$ .

In this case the right hand and left hand sides are equal to only possible at $x$ = $-1$ .

Hence mod of $x$ = 1.