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

The solutions to the equation z^2+4z=2 are z=-2+sqrt(6),z=-2-sqrt(6)

Find the zeros of z^2+4z=2

The zeros of z^2+4z=2 are z=-2+sqrt(6),z=-2-sqrt(6)

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z^{2} + 4 z = 2

z = - 2 + 6, z = - 2 - 6

Decimal Notation

z = 0.44948 \dots, z = - 4.44948 \dots

Solution steps

z^{2} + 4 z = 2

Move

2

to the left side

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

Solve with the quadratic formula

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

4^{2} - 4 \cdot 1 \cdot (- 2) = 26

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

Separate the solutions

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

z = \frac{- 4 + 2 6}{2 \cdot 1} : - 2 + 6

z = \frac{- 4 - 2 6}{2 \cdot 1} : - 2 - 6

The solutions to the quadratic equation are:

z = - 2 + 6, z = - 2 - 6

- n + 2 - n < - \frac{2}{3} (9 n - 63)

5 x^{2} - 2 x + 3 = 0

500 x^{3} + 108 y^{18}

- 3 (5 - 9 v) = 25 + 7 v

∣ 3 x ∣ < 6

What are the solutions to the equation z^2+4z=2 ?
The solutions to the equation z^2+4z=2 are z=-2+sqrt(6),z=-2-sqrt(6)
Find the zeros of z^2+4z=2
The zeros of z^2+4z=2 are z=-2+sqrt(6),z=-2-sqrt(6)