What is a^2+5a-95/15 =b ?

The solution to a^2+5a-95/15 =b is Parabola with (h,k)=(-5/2 ,-151/12),p= 1/4

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a^2+5a-95/15 =b

a^{2} + 5 a - \frac{95}{15} = b

(h, k) = (- \frac{5}{2}, - \frac{151}{12}), p = \frac{1}{4}

Solution steps

a^{2} + 5 a - \frac{95}{15} = b

Rewrite

a^{2} + 5 a - \frac{95}{15} = b

in the standard form

4 \cdot \frac{1}{4} (b - (- \frac{151}{12})) = (a - (- \frac{5}{2}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{5}{2}, - \frac{151}{12}), p = \frac{1}{4}

\frac{( y - 2 ) ^{2}}{3} = 8 - x

(y - 1)^{2} = - 8 (x + 4)

X \leq (Y + 2)^{2}

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

12 x^{2} + 10 x - 6 y + 1 = 0

What is a^2+5a-95/15 =b ?
The solution to a^2+5a-95/15 =b is Parabola with (h,k)=(-5/2 ,-151/12),p= 1/4