What are the solutions to the equation 3z^2+11z-4=0 ?

The solutions to the equation 3z^2+11z-4=0 are z= 1/3 ,z=-4

Find the zeros of 3z^2+11z-4=0

The zeros of 3z^2+11z-4=0 are z= 1/3 ,z=-4

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3 z^{2} + 11 z - 4 = 0

z = \frac{1}{3}, z = - 4

Decimal Notation

z = 0.33333 \dots, z = - 4

Solution steps

3 z^{2} + 11 z - 4 = 0

Solve with the quadratic formula

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

1 1^{2} - 4 \cdot 3 (- 4) = 13

z_{1, 2} = \frac{- 11 \pm 13}{2 \cdot 3}

Separate the solutions

z_{1} = \frac{- 11 + 13}{2 \cdot 3}, z_{2} = \frac{- 11 - 13}{2 \cdot 3}

z = \frac{- 11 + 13}{2 \cdot 3} : \frac{1}{3}

z = \frac{- 11 - 13}{2 \cdot 3} : - 4

The solutions to the quadratic equation are:

z = \frac{1}{3}, z = - 4

2 c^{2} - 9 c + 9

\frac{3}{2}

78 x^{2} + 47 x - 41 = 0

9 x^{2} - 6 x - 3 x + 2

- 2 x + 3 \leq 5

What are the solutions to the equation 3z^2+11z-4=0 ?
The solutions to the equation 3z^2+11z-4=0 are z= 1/3 ,z=-4
Find the zeros of 3z^2+11z-4=0
The zeros of 3z^2+11z-4=0 are z= 1/3 ,z=-4