Let x2+34x−71x2+2x−7=y,x∈R
x2+34x−71=y(x2+2x−7)
x2+34x−71=yx2+2yx−7y
(y−1)x2+(2y−34)x+71−7y=0
As x∈R discriminant must be greater than zero
(2y−34)2≥4(y−1)(71−7y)
(y−17)2≥(y−1)(71−7y)
y2+289−34y≥71y−71−7y2+7y
8y2−112y+360≥0
y2−14y+45≥0
y2−5y−9y+45≥0
y(y−5)−9(y−5)≥0
(y−5)(y−9)≥0
So y∈(−∞,5]∪[9,∞)
So it can be in (5,9)