What is 2x^2-7x+K(x)=0 ?

The solution to 2x^2-7x+K(x)=0 is Parabola with (h,k)=(7/4 , 49/8),p=-1/8

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2x^2-7x+K(x)=0

2 x^{2} - 7 x + K (x) = 0

(h, k) = (\frac{7}{4}, \frac{49}{8}), p = - \frac{1}{8}

Solution steps

2 x^{2} - 7 x + y = 0

Rewrite

2 x^{2} - 7 x + y = 0

in the standard form

4 (- \frac{1}{8}) (y - \frac{49}{8}) = (x - \frac{7}{4})^{2}

Therefore parabola properties are:

(h, k) = (\frac{7}{4}, \frac{49}{8}), p = - \frac{1}{8}

9 x^{2} = x + y

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

x = y^{2} + 14

2 y^{2} - 2 x - 2 y + 9 = 0

\frac{( z ^{2} + 2 \cdot z + I )}{( 1 - z )} = - 1

What is 2x^2-7x+K(x)=0 ?
The solution to 2x^2-7x+K(x)=0 is Parabola with (h,k)=(7/4 , 49/8),p=-1/8