What are the solutions to the equation z^2-7z+13=0 ?

The solutions to the equation z^2-7z+13=0 are z= 7/2+i(sqrt(3))/2 ,z= 7/2-i(sqrt(3))/2

Find the zeros of z^2-7z+13=0

The zeros of z^2-7z+13=0 are z= 7/2+i(sqrt(3))/2 ,z= 7/2-i(sqrt(3))/2

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z^{2} - 7 z + 13 = 0

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

Solution steps

z^{2} - 7 z + 13 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot 13}{2 \cdot 1}

Simplify

(- 7)^{2} - 4 \cdot 1 \cdot 13 : 3 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

m x^{2} - 5 x + 2 = n x - 1

(2 x - 1) (x + 3) = 4

x^{2} - \frac{2}{3} x = \frac{3}{5}

3 x^{2} + 7 = 4

x^{2} = a, a > 0

What are the solutions to the equation z^2-7z+13=0 ?
The solutions to the equation z^2-7z+13=0 are z= 7/2+i(sqrt(3))/2 ,z= 7/2-i(sqrt(3))/2
Find the zeros of z^2-7z+13=0
The zeros of z^2-7z+13=0 are z= 7/2+i(sqrt(3))/2 ,z= 7/2-i(sqrt(3))/2