What are the solutions to the equation z^2+4z-12=0 ?

The solutions to the equation z^2+4z-12=0 are z=2,z=-6

Find the zeros of z^2+4z-12=0

The zeros of z^2+4z-12=0 are z=2,z=-6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

z^{2} + 4 z - 12 = 0

z = 2, z = - 6

Solution steps

z^{2} + 4 z - 12 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 12 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 12) = 8

z_{1, 2} = \frac{- 4 \pm 8}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- 4 + 8}{2 \cdot 1}, z_{2} = \frac{- 4 - 8}{2 \cdot 1}

z = \frac{- 4 + 8}{2 \cdot 1} : 2

z = \frac{- 4 - 8}{2 \cdot 1} : - 6

The solutions to the quadratic equation are:

z = 2, z = - 6

203

2 x^{2} = 19 x + 33

- \frac{4}{7} x = - 1

(310)^{2}

What are the solutions to the equation z^2+4z-12=0 ?
The solutions to the equation z^2+4z-12=0 are z=2,z=-6
Find the zeros of z^2+4z-12=0
The zeros of z^2+4z-12=0 are z=2,z=-6