What is x^2-5x+(m(x))/4 =0 ?

The solution to x^2-5x+(m(x))/4 =0 is Parabola with (h,k)=(5/2 ,25),p=-1/16

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x^2-5x+(m(x))/4 =0

x^{2} - 5 x + \frac{m ( x )}{4} = 0

(h, k) = (\frac{5}{2}, 25), p = - \frac{1}{16}

Solution steps

x^{2} - 5 x + \frac{y}{4} = 0

Rewrite

x^{2} - 5 x + \frac{y}{4} = 0

in the standard form

4 (- \frac{1}{16}) (y - 25) = (x - \frac{5}{2})^{2}

Therefore parabola properties are:

(h, k) = (\frac{5}{2}, 25), p = - \frac{1}{16}

(y - 2)^{2} = 8 (x - 4)

2 x^{2} + 8 y - 10 = 0

2 x^{2} - 4 x - 4 y + 18 = 0

x^{2} + 14 x - y - 14 = 0

x^{2} + 2 y = 10

What is x^2-5x+(m(x))/4 =0 ?
The solution to x^2-5x+(m(x))/4 =0 is Parabola with (h,k)=(5/2 ,25),p=-1/16