What are the solutions to the equation 6z^2-11z=-7z-4 ?

The solutions to the equation 6z^2-11z=-7z-4 are z= 1/3+i(sqrt(5))/3 ,z= 1/3-i(sqrt(5))/3

Find the zeros of 6z^2-11z=-7z-4

The zeros of 6z^2-11z=-7z-4 are z= 1/3+i(sqrt(5))/3 ,z= 1/3-i(sqrt(5))/3

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6 z^{2} - 11 z = - 7 z - 4

z = \frac{1}{3} + i \frac{5}{3}, z = \frac{1}{3} - i \frac{5}{3}

Solution steps

6 z^{2} - 11 z = - 7 z - 4

Move

4

to the left side

6 z^{2} - 11 z + 4 = - 7 z

Move

7 z

to the left side

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

Solve with the quadratic formula

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

Simplify

(- 4)^{2} - 4 \cdot 6 \cdot 4 : 45 i

z_{1, 2} = \frac{- ( - 4 ) \pm 4 5 i}{2 \cdot 6}

Separate the solutions

z_{1} = \frac{- ( - 4 ) + 4 5 i}{2 \cdot 6}, z_{2} = \frac{- ( - 4 ) - 4 5 i}{2 \cdot 6}

z = \frac{- ( - 4 ) + 4 5 i}{2 \cdot 6} : \frac{1}{3} + i \frac{5}{3}

z = \frac{- ( - 4 ) - 4 5 i}{2 \cdot 6} : \frac{1}{3} - i \frac{5}{3}

The solutions to the quadratic equation are:

z = \frac{1}{3} + i \frac{5}{3}, z = \frac{1}{3} - i \frac{5}{3}

3 x + 1 < 2 x - 2

x^{4} - 145 x^{2} + 144 \geq 0

∣ 2 x + 8 ∣ = 12

4 x^{2} + 20 x + 25 = 49

x^{2} + 4 x + 4 = 12

What are the solutions to the equation 6z^2-11z=-7z-4 ?
The solutions to the equation 6z^2-11z=-7z-4 are z= 1/3+i(sqrt(5))/3 ,z= 1/3-i(sqrt(5))/3
Find the zeros of 6z^2-11z=-7z-4
The zeros of 6z^2-11z=-7z-4 are z= 1/3+i(sqrt(5))/3 ,z= 1/3-i(sqrt(5))/3