What is C^2+3C+5T=1300 ?

The solution to C^2+3C+5T=1300 is Parabola with (h,k)=(-3/2 , 5209/20),p=-5/4

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C^2+3C+5T=1300

C^{2} + 3 C + 5 T = 1300

(h, k) = (- \frac{3}{2}, \frac{5209}{20}), p = - \frac{5}{4}

Solution steps

C^{2} + 3 C + 5 T = 1300

Rewrite

C^{2} + 3 C + 5 T = 1300

in the standard form

4 (- \frac{5}{4}) (T - \frac{5209}{20}) = (C - (- \frac{3}{2}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{3}{2}, \frac{5209}{20}), p = - \frac{5}{4}

x^{2} = 0.1 y

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

2 x + y = x^{2}

y^{2} + 2 y - 4 x = 7

x^{2} = - 7 y

What is C^2+3C+5T=1300 ?
The solution to C^2+3C+5T=1300 is Parabola with (h,k)=(-3/2 , 5209/20),p=-5/4