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

The solutions to the equation z^2-2z=-5 are z=1+2i,z=1-2i

Find the zeros of z^2-2z=-5

The zeros of z^2-2z=-5 are z=1+2i,z=1-2i

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z^{2} - 2 z = - 5

z = 1 + 2 i, z = 1 - 2 i

Solution steps

z^{2} - 2 z = - 5

Move

5

to the left side

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

Solve with the quadratic formula

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

Simplify

(- 2)^{2} - 4 \cdot 1 \cdot 5 : 4 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

z = 1 + 2 i, z = 1 - 2 i

x^{2} = 0.0081

x, 3 x 7 - 4 x^{2} y + x y^{3} = 25

x, x^{2} - 2 (1 + 3 p) x + 7 (3 + 2 p) = 0

lo g_{10} (9) (x^{2} - 4 x) = lo g_{10} (9) (3 x - 10)

3 n^{2} + 4 n + 4 = 0

What are the solutions to the equation z^2-2z=-5 ?
The solutions to the equation z^2-2z=-5 are z=1+2i,z=1-2i
Find the zeros of z^2-2z=-5
The zeros of z^2-2z=-5 are z=1+2i,z=1-2i