What are the solutions to the equation -z^2+3z-5=0 ?

The solutions to the equation -z^2+3z-5=0 are z= 3/2-i(sqrt(11))/2 ,z= 3/2+i(sqrt(11))/2

Find the zeros of -z^2+3z-5=0

The zeros of -z^2+3z-5=0 are z= 3/2-i(sqrt(11))/2 ,z= 3/2+i(sqrt(11))/2

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- z^{2} + 3 z - 5 = 0

z = \frac{3}{2} - i \frac{11}{2}, z = \frac{3}{2} + i \frac{11}{2}

Solution steps

- z^{2} + 3 z - 5 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 1 ) ( - 5 )}{2 ( - 1 )}

Simplify

3^{2} - 4 (- 1) (- 5) : 11 i

z_{1, 2} = \frac{- 3 \pm 11 i}{2 ( - 1 )}

Separate the solutions

z_{1} = \frac{- 3 + 11 i}{2 ( - 1 )}, z_{2} = \frac{- 3 - 11 i}{2 ( - 1 )}

z = \frac{- 3 + 11 i}{2 ( - 1 )} : \frac{3}{2} - i \frac{11}{2}

z = \frac{- 3 - 11 i}{2 ( - 1 )} : \frac{3}{2} + i \frac{11}{2}

The solutions to the quadratic equation are:

z = \frac{3}{2} - i \frac{11}{2}, z = \frac{3}{2} + i \frac{11}{2}

512

\frac{2}{3} \cdot \frac{2}{5}

c^{2} - 10 c + 25

\frac{3}{4} (x - 1) + x = 1

125 - 8 x^{3}

What are the solutions to the equation -z^2+3z-5=0 ?
The solutions to the equation -z^2+3z-5=0 are z= 3/2-i(sqrt(11))/2 ,z= 3/2+i(sqrt(11))/2
Find the zeros of -z^2+3z-5=0
The zeros of -z^2+3z-5=0 are z= 3/2-i(sqrt(11))/2 ,z= 3/2+i(sqrt(11))/2