What are the solutions to the equation z^2-4z+13=0 ?

The solutions to the equation z^2-4z+13=0 are z=2+3i,z=2-3i

Find the zeros of z^2-4z+13=0

The zeros of z^2-4z+13=0 are z=2+3i,z=2-3i

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z^{2} - 4 z + 13 = 0

z = 2 + 3 i, z = 2 - 3 i

Solution steps

z^{2} - 4 z + 13 = 0

Solve with the quadratic formula

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

Simplify

(- 4)^{2} - 4 \cdot 1 \cdot 13 : 6 i

z_{1, 2} = \frac{- ( - 4 ) \pm 6 i}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- ( - 4 ) + 6 i}{2 \cdot 1}, z_{2} = \frac{- ( - 4 ) - 6 i}{2 \cdot 1}

z = \frac{- ( - 4 ) + 6 i}{2 \cdot 1} : 2 + 3 i

z = \frac{- ( - 4 ) - 6 i}{2 \cdot 1} : 2 - 3 i

The solutions to the quadratic equation are:

z = 2 + 3 i, z = 2 - 3 i

8 x^{2} - 2 x = 7

x^{2} - 11 x - 5 = 0

m^{2} - 25 = 0

3 y^{2} - 21 y = 0

3 x^{2} - 17 x + 20 = 0

What are the solutions to the equation z^2-4z+13=0 ?
The solutions to the equation z^2-4z+13=0 are z=2+3i,z=2-3i
Find the zeros of z^2-4z+13=0
The zeros of z^2-4z+13=0 are z=2+3i,z=2-3i