What is 7x^2+3x+h(x)=0 ?

The solution to 7x^2+3x+h(x)=0 is Parabola with (h,k)=(-3/14 , 9/28),p=-1/28

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7x^2+3x+h(x)=0

7 x^{2} + 3 x + h (x) = 0

(h, k) = (- \frac{3}{14}, \frac{9}{28}), p = - \frac{1}{28}

Solution steps

7 x^{2} + 3 x + y = 0

Rewrite

7 x^{2} + 3 x + y = 0

in the standard form

4 (- \frac{1}{28}) (y - \frac{9}{28}) = (x - (- \frac{3}{14}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{3}{14}, \frac{9}{28}), p = - \frac{1}{28}

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

x, 2 m (x)^{2} + 6 x = 0

y^{2} + 12 y = - x + 1

y = 6 x - 2 x^{2}

(x - 9)^{2} + (y)^{2} = (x - 6^{2}) + (y + 3)^{2}

What is 7x^2+3x+h(x)=0 ?
The solution to 7x^2+3x+h(x)=0 is Parabola with (h,k)=(-3/14 , 9/28),p=-1/28