What is 8x^2+3x+h(x)=0 ?

The solution to 8x^2+3x+h(x)=0 is Parabola with (h,k)=(-3/16 , 9/32),p=-1/32

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8x^2+3x+h(x)=0

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

(h, k) = (- \frac{3}{16}, \frac{9}{32}), p = - \frac{1}{32}

Solution steps

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

Rewrite

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

in the standard form

4 (- \frac{1}{32}) (y - \frac{9}{32}) = (x - (- \frac{3}{16}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{3}{16}, \frac{9}{32}), p = - \frac{1}{32}

16 x + 2 y^{2} = 0

t, x = 7 \cdot tan (7 4^{\circ}) f t

Y (x) > x^{2} - 2

(2 x - 10) + (2 x - 3) + (x - 2) = 50 f (x, y) y

x - y^{2} = - 7

What is 8x^2+3x+h(x)=0 ?
The solution to 8x^2+3x+h(x)=0 is Parabola with (h,k)=(-3/16 , 9/32),p=-1/32