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

The solutions to the equation z^2-2z=3 are z=3,z=-1

Find the zeros of z^2-2z=3

The zeros of z^2-2z=3 are z=3,z=-1

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z^{2} - 2 z = 3

z = 3, z = - 1

Solution steps

z^{2} - 2 z = 3

Move

3

to the left side

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 1 \cdot (- 3) = 4

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

z = 3, z = - 1

3 x^{2} - 6 x - 7 = x^{2} + 3 x - x (x + 1) + 3

x, 3 x^{2} - 3 y^{2} = 0

2 x^{2} - 2 x = x^{2} + 2 x + 1

3 (x + 2)^{2} + 4 = 40

3 x (2 x + 9) = 10 x

What are the solutions to the equation z^2-2z=3 ?
The solutions to the equation z^2-2z=3 are z=3,z=-1
Find the zeros of z^2-2z=3
The zeros of z^2-2z=3 are z=3,z=-1