What are the solutions to the equation z^2+4z=-3 ?

The solutions to the equation z^2+4z=-3 are z=-1,z=-3

Find the zeros of z^2+4z=-3

The zeros of z^2+4z=-3 are z=-1,z=-3

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z^{2} + 4 z = - 3

z = - 1, z = - 3

Solution steps

z^{2} + 4 z = - 3

Move

3

to the left side

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

z = - 1, z = - 3

340

\frac{3}{2} - \frac{1}{1}

- 81 = 9 x

x^{4} - x^{2} - 30

x^{3} - 12 x^{2} + 41 x - 30

What are the solutions to the equation z^2+4z=-3 ?
The solutions to the equation z^2+4z=-3 are z=-1,z=-3
Find the zeros of z^2+4z=-3
The zeros of z^2+4z=-3 are z=-1,z=-3