What are the solutions to the equation z^2-3z+4=0 ?

The solutions to the equation z^2-3z+4=0 are z= 3/2+i(sqrt(7))/2 ,z= 3/2-i(sqrt(7))/2

Find the zeros of z^2-3z+4=0

The zeros of z^2-3z+4=0 are z= 3/2+i(sqrt(7))/2 ,z= 3/2-i(sqrt(7))/2

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z^{2} - 3 z + 4 = 0

z = \frac{3}{2} + i \frac{7}{2}, z = \frac{3}{2} - i \frac{7}{2}

Solution steps

z^{2} - 3 z + 4 = 0

Solve with the quadratic formula

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

Simplify

(- 3)^{2} - 4 \cdot 1 \cdot 4 : 7 i

z_{1, 2} = \frac{- ( - 3 ) \pm 7 i}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- ( - 3 ) + 7 i}{2 \cdot 1}, z_{2} = \frac{- ( - 3 ) - 7 i}{2 \cdot 1}

z = \frac{- ( - 3 ) + 7 i}{2 \cdot 1} : \frac{3}{2} + i \frac{7}{2}

z = \frac{- ( - 3 ) - 7 i}{2 \cdot 1} : \frac{3}{2} - i \frac{7}{2}

The solutions to the quadratic equation are:

z = \frac{3}{2} + i \frac{7}{2}, z = \frac{3}{2} - i \frac{7}{2}

y, 3 x^{2} + 2 x y + 5 y^{2} = 24

\frac{4}{3} x^{2} - 6 = - 70

5 (2 x + 9)^{2} - 21 = 4

9 b^{2} + 12 b + 5 = 1

p^{2} + 3 = - 10

What are the solutions to the equation z^2-3z+4=0 ?
The solutions to the equation z^2-3z+4=0 are z= 3/2+i(sqrt(7))/2 ,z= 3/2-i(sqrt(7))/2
Find the zeros of z^2-3z+4=0
The zeros of z^2-3z+4=0 are z= 3/2+i(sqrt(7))/2 ,z= 3/2-i(sqrt(7))/2