What is 17-x+v169-x^2>0 ?

The solution to 17-x+v169-x^2>0 is Parabola with (h,k)=(-1/2 ,-69/676),p= 169/4

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17-x+v169-x^2>0

17 - x + v 169 - x^{2} > 0

(h, k) = (- \frac{1}{2}, - \frac{69}{676}), p = \frac{169}{4}

Solution steps

17 - x + v \cdot 169 - x^{2} > 0

Rewrite

17 - x + v \cdot 169 - x^{2} > 0

in the standard form

4 \cdot \frac{169}{4} (v - (- \frac{69}{676})) = (x - (- \frac{1}{2}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{1}{2}, - \frac{69}{676}), p = \frac{169}{4}

y = - \frac{5}{y} x - 1

x^{2} - 6 y - 39 = 0

7 x^{2} + 9 x + k (x) = 0

x, Y (x)^{2} = 8 x

x^{2} + 8 y + 80 = 0

What is 17-x+v169-x^2>0 ?
The solution to 17-x+v169-x^2>0 is Parabola with (h,k)=(-1/2 ,-69/676),p= 169/4