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=sqrt(6)-2

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

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

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

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

Decimal Notation

z = - 4.44948 \dots, z = 0.44948 \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 ( - 1 ) \cdot 2}{2 ( - 1 )}

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

\frac{6 x y ^{\frac{3}{4}}}{3 x ^{\frac{1}{2}} y ^{\frac{1}{2}}}

a^{6} b^{7} - a^{3} b^{2} + a^{2} b^{4} - a^{2} b^{2}

216 x^{3} + 1

\frac{12 x y ^{5}}{20 x ^{3} y ^{7}}

\frac{a + 7}{a ^{2} + 6 a - 7}

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