What is the answer to i=-24t^2+333t-32 ?

The answer to i=-24t^2+333t-32 is t=0.09677…+0.00304…i,t=13.77822…-0.00304…i

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i=-24t^2+333t-32

i = - 24 t^{2} + 333 t - 32

t = 0.09677 \dots + 0.00304 \dots i, t = 13.77822 \dots - 0.00304 \dots i

Solution steps

i = - 24 t^{2} + 333 t - 32

Substitute

t = x + y i

i = - 24 (x + y i)^{2} + 333 (x + y i) - 32

Expand

- 24 (x + y i)^{2} + 333 (x + y i) - 32 : (- 24 x^{2} + 24 y^{2} + 333 x - 32) + i (333 y - 48 x y)

i = (- 24 x^{2} + 24 y^{2} + 333 x - 32) + i (333 y - 48 x y)

Rewrite

i

in standard complex form:

0 + i

0 + i = (- 24 x^{2} + 24 y^{2} + 333 x - 32) + i (333 y - 48 x y)

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

[0 = - 24 x^{2} + 24 y^{2} + 333 x - 32 1 = 333 y - 48 x y]

[0 = - 24 x^{2} + 24 y^{2} + 333 x - 32 1 = 333 y - 48 x y] : (x = 0.09677 \dots, x = 13.77822 \dots, y = 0.00304 \dots y = - 0.00304 \dots)

Substitute back

t = x + y i

t = 0.09677 \dots + 0.00304 \dots i, t = 13.77822 \dots - 0.00304 \dots i

(5 - \frac{1}{2} x) + (y - 1) i = 7 - 2 i

(x + 2 i y)^{2} - 8 i + 15 = 0

x = \frac{y}{( 1 + i \cdot n )}

(2 + y i) (x + i) = 1 + 3 i

z^{2} - (4 + 2 i) z + 2 + 4 i = 0

What is the answer to i=-24t^2+333t-32 ?
The answer to i=-24t^2+333t-32 is t=0.09677…+0.00304…i,t=13.77822…-0.00304…i