What is x^2+(3-i)z+2-6i=0 ?

The solution to x^2+(3-i)z+2-6i=0 is Parabola with (h,k)=(0,-6/5+i 8/5),p=-3/4+i 1/4

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x^2+(3-i)z+2-6i=0

x^{2} + (3 - i) z + 2 - 6 i = 0

(h, k) = (0, - \frac{6}{5} + i \frac{8}{5}), p = - \frac{3}{4} + i \frac{1}{4}

Solution steps

x^{2} + (3 - i) z + 2 - 6 i = 0

Rewrite

x^{2} + (3 - i) z + 2 - 6 i = 0

in the standard form:

4 (- \frac{3}{4} + i \frac{1}{4}) (z - (- \frac{6}{5} + i \frac{8}{5})) = (x - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, - \frac{6}{5} + i \frac{8}{5}), p = - \frac{3}{4} + i \frac{1}{4}

y - 1 = 2 (x - 3)^{2}

x^{2} + x - 12 = i z a r

Y (x) + 1 = (x - 1)^{2}

x^{2} = 13 y

x^{2} + 2 \cdot 71 = o

What is x^2+(3-i)z+2-6i=0 ?
The solution to x^2+(3-i)z+2-6i=0 is Parabola with (h,k)=(0,-6/5+i 8/5),p=-3/4+i 1/4