What is (10+y)2-y^2=-3x+12 ?

The solution to (10+y)2-y^2=-3x+12 is Parabola with (h,k)=(-3,1),p= 3/4

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(10+y)2-y^2=-3x+12

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

(h, k) = (- 3, 1), p = \frac{3}{4}

Solution steps

(10 + y) \cdot 2 - y^{2} = - 3 x + 12

Rewrite

(10 + y) \cdot 2 - y^{2} = - 3 x + 12

in the standard form

4 \cdot \frac{3}{4} (x - (- 3)) = (y - 1)^{2}

Therefore parabola properties are:

(h, k) = (- 3, 1), p = \frac{3}{4}

(y - 3)^{2} = - 20 (x + 2)

3 a - 5 (b - a) = (a - b) \cdot (a + b) - a (a - 1)

0 = x^{2} + 14 x - 4 y + 48

8 x^{2} + 3 y - 12 x + 6 y + 18 = 0

x^{2} + 6 x + 4 y + 8 = 0

What is (10+y)2-y^2=-3x+12 ?
The solution to (10+y)2-y^2=-3x+12 is Parabola with (h,k)=(-3,1),p= 3/4