What are the solutions to the equation 5z^2+19z-4=0 ?

The solutions to the equation 5z^2+19z-4=0 are z= 1/5 ,z=-4

Find the zeros of 5z^2+19z-4=0

The zeros of 5z^2+19z-4=0 are z= 1/5 ,z=-4

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5 z^{2} + 19 z - 4 = 0

z = \frac{1}{5}, z = - 4

Decimal Notation

z = 0.2, z = - 4

Solution steps

5 z^{2} + 19 z - 4 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- 19 \pm 1 9 ^{2} - 4 \cdot 5 ( - 4 )}{2 \cdot 5}

1 9^{2} - 4 \cdot 5 (- 4) = 21

z_{1, 2} = \frac{- 19 \pm 21}{2 \cdot 5}

Separate the solutions

z_{1} = \frac{- 19 + 21}{2 \cdot 5}, z_{2} = \frac{- 19 - 21}{2 \cdot 5}

z = \frac{- 19 + 21}{2 \cdot 5} : \frac{1}{5}

z = \frac{- 19 - 21}{2 \cdot 5} : - 4

The solutions to the quadratic equation are:

z = \frac{1}{5}, z = - 4

\frac{x ^{2} - 9}{x + 2} < 0

64 x^{3} + 1

x - (2 x - 1) \leq 3 (x - 4) + 1

\frac{3}{4 i}

\frac{x + 3}{2} - 2 > 4

What are the solutions to the equation 5z^2+19z-4=0 ?
The solutions to the equation 5z^2+19z-4=0 are z= 1/5 ,z=-4
Find the zeros of 5z^2+19z-4=0
The zeros of 5z^2+19z-4=0 are z= 1/5 ,z=-4