What are the solutions to the equation z^2-10z+29=0 ?

The solutions to the equation z^2-10z+29=0 are z=5+2i,z=5-2i

Find the zeros of z^2-10z+29=0

The zeros of z^2-10z+29=0 are z=5+2i,z=5-2i

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z^{2} - 10 z + 29 = 0

z = 5 + 2 i, z = 5 - 2 i

Solution steps

z^{2} - 10 z + 29 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 1 \cdot 29}{2 \cdot 1}

Simplify

(- 10)^{2} - 4 \cdot 1 \cdot 29 : 4 i

z_{1, 2} = \frac{- ( - 10 ) \pm 4 i}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- ( - 10 ) + 4 i}{2 \cdot 1}, z_{2} = \frac{- ( - 10 ) - 4 i}{2 \cdot 1}

z = \frac{- ( - 10 ) + 4 i}{2 \cdot 1} : 5 + 2 i

z = \frac{- ( - 10 ) - 4 i}{2 \cdot 1} : 5 - 2 i

The solutions to the quadratic equation are:

z = 5 + 2 i, z = 5 - 2 i

x (x + 2) = 4

x^{2} - 6 = - 2 x

4 5^{2} + 2 4^{2} = x^{2}

4 x^{2} - 5 x^{2} - 20 = 0

x^{2} - 7 x = - \frac{24}{4}

What are the solutions to the equation z^2-10z+29=0 ?
The solutions to the equation z^2-10z+29=0 are z=5+2i,z=5-2i
Find the zeros of z^2-10z+29=0
The zeros of z^2-10z+29=0 are z=5+2i,z=5-2i