What is -(-2)x^2+12x+15=-3x^2+4y ?

The solution to -(-2)x^2+12x+15=-3x^2+4y is Parabola with (h,k)=(-6/5 , 39/20),p= 1/5

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-(-2)x^2+12x+15=-3x^2+4y

- (- 2) x^{2} + 12 x + 15 = - 3 x^{2} + 4 y

(h, k) = (- \frac{6}{5}, \frac{39}{20}), p = \frac{1}{5}

Solution steps

- (- 2) x^{2} + 12 x + 15 = - 3 x^{2} + 4 y

Rewrite

- (- 2) x^{2} + 12 x + 15 = - 3 x^{2} + 4 y

in the standard form

4 \cdot \frac{1}{5} (y - \frac{39}{20}) = (x - (- \frac{6}{5}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{6}{5}, \frac{39}{20}), p = \frac{1}{5}

x^{2} + 12 x + k (x) = 0

3 y^{2} - 4 x - 6 y + 23 = 0

y^{2} + 10 y + 2 x + 11 = 0

(y - 1)^{2} = - 16 (x - 1)

3 x^{2} - 12 x + q (x) = 0

What is -(-2)x^2+12x+15=-3x^2+4y ?
The solution to -(-2)x^2+12x+15=-3x^2+4y is Parabola with (h,k)=(-6/5 , 39/20),p= 1/5