What are the solutions to the equation z(3z+11)=20 ?

The solutions to the equation z(3z+11)=20 are z= 4/3 ,z=-5

Find the zeros of z(3z+11)=20

The zeros of z(3z+11)=20 are z= 4/3 ,z=-5

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z (3 z + 11) = 20

z = \frac{4}{3}, z = - 5

Decimal Notation

z = 1.33333 \dots, z = - 5

Solution steps

z (3 z + 11) = 20

Expand

z (3 z + 11) : 3 z^{2} + 11 z

3 z^{2} + 11 z = 20

Move

20

to the left side

3 z^{2} + 11 z - 20 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 3 ( - 20 )}{2 \cdot 3}

1 1^{2} - 4 \cdot 3 (- 20) = 19

z_{1, 2} = \frac{- 11 \pm 19}{2 \cdot 3}

Separate the solutions

z_{1} = \frac{- 11 + 19}{2 \cdot 3}, z_{2} = \frac{- 11 - 19}{2 \cdot 3}

z = \frac{- 11 + 19}{2 \cdot 3} : \frac{4}{3}

z = \frac{- 11 - 19}{2 \cdot 3} : - 5

The solutions to the quadratic equation are:

z = \frac{4}{3}, z = - 5

4 x^{2} = 405 - x^{2}

- 4 x^{2} + 5 = 0

x^{2} - 18 + 81 = 0

- 4 x^{2} + 13 = 0

x^{2} - 7 = 18

What are the solutions to the equation z(3z+11)=20 ?
The solutions to the equation z(3z+11)=20 are z= 4/3 ,z=-5
Find the zeros of z(3z+11)=20
The zeros of z(3z+11)=20 are z= 4/3 ,z=-5