What is the answer to 4z^2+(-13+18i)z-5-10i=0 ?

The answer to 4z^2+(-13+18i)z-5-10i=0 is z=3-4i,z= 1/4-1/2 i

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4z^2+(-13+18i)z-5-10i=0

4 z^{2} + (- 13 + 18 i) z - 5 - 10 i = 0

z = 3 - 4 i, z = \frac{1}{4} - \frac{1}{2} i

Solution steps

4 z^{2} + (- 13 + 18 i) z - 5 - 10 i = 0

Substitute

z = x + y i

4 (x + y i)^{2} + (- 13 + 18 i) (x + y i) - 5 - 10 i = 0

Expand

4 (x + y i)^{2} + (- 13 + 18 i) (x + y i) - 5 - 10 i : (4 x^{2} - 4 y^{2} - 13 x - 18 y - 5) + i (- 10 - 13 y + 18 x + 8 x y)

(4 x^{2} - 4 y^{2} - 13 x - 18 y - 5) + i (- 10 - 13 y + 18 x + 8 x y) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(4 x^{2} - 4 y^{2} - 13 x - 18 y - 5) + i (- 10 - 13 y + 18 x + 8 x y) = 0 + 0 i

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[4 x^{2} - 4 y^{2} - 13 x - 18 y - 5 = 0 - 10 - 13 y + 18 x + 8 x y = 0]

[4 x^{2} - 4 y^{2} - 13 x - 18 y - 5 = 0 - 10 - 13 y + 18 x + 8 x y = 0] : (x = 3, x = \frac{1}{4}, y = - 4 y = - \frac{1}{2})

Substitute back

z = x + y i

z = 3 - 4 i, z = \frac{1}{4} - \frac{1}{2} i

x^{2} + 4 - 3 i = 0

- 3 ai - (- 1 - i) b = 3 a - 2

i (x) = \frac{1}{x}

x^{2} = - i

z^{2} + 2 z + 1 + i = 0

What is the answer to 4z^2+(-13+18i)z-5-10i=0 ?
The answer to 4z^2+(-13+18i)z-5-10i=0 is z=3-4i,z= 1/4-1/2 i