What is x^2+3x+5y-5=0 ?

The solution to x^2+3x+5y-5=0 is Parabola with (h,k)=(-3/2 , 29/20),p=-5/4

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x^2+3x+5y-5=0

x^{2} + 3 x + 5 y - 5 = 0

(h, k) = (- \frac{3}{2}, \frac{29}{20}), p = - \frac{5}{4}

Solution steps

x^{2} + 3 x + 5 y - 5 = 0

Rewrite

x^{2} + 3 x + 5 y - 5 = 0

in the standard form

4 (- \frac{5}{4}) (y - \frac{29}{20}) = (x - (- \frac{3}{2}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{3}{2}, \frac{29}{20}), p = - \frac{5}{4}

y^{2} = 4 c (x + c)

y = - (x - 2)^{2} - 6

y^{2} + 2 y - x = 0

(y - 4)^{2} = (x - 3)

y = 2 (x - 5)^{2} + 3

What is x^2+3x+5y-5=0 ?
The solution to x^2+3x+5y-5=0 is Parabola with (h,k)=(-3/2 , 29/20),p=-5/4