What are the solutions to the equation 19j^2+31j=0 ?

The solutions to the equation 19j^2+31j=0 are j=0,j=-31/19

Find the zeros of 19j^2+31j=0

The zeros of 19j^2+31j=0 are j=0,j=-31/19

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19 j^{2} + 31 j = 0

j = 0, j = - \frac{31}{19}

Decimal Notation

j = 0, j = - 1.63157 \dots

Solution steps

19 j^{2} + 31 j = 0

Solve with the quadratic formula

j_{1, 2} = \frac{- 31 \pm 3 1 ^{2} - 4 \cdot 19 \cdot 0}{2 \cdot 19}

3 1^{2} - 4 \cdot 19 \cdot 0 = 31

j_{1, 2} = \frac{- 31 \pm 31}{2 \cdot 19}

Separate the solutions

j_{1} = \frac{- 31 + 31}{2 \cdot 19}, j_{2} = \frac{- 31 - 31}{2 \cdot 19}

j = \frac{- 31 + 31}{2 \cdot 19} : 0

j = \frac{- 31 - 31}{2 \cdot 19} : - \frac{31}{19}

The solutions to the quadratic equation are:

j = 0, j = - \frac{31}{19}

∣ u - 6 ∣ = - \frac{1}{5}

- 5 = 52

1 0^{5} \cdot \frac{1 0 ^{4}}{1 0 ^{2}}

7 x - 14 = - 28

\frac{( 265 \cdot 1 0 ^{- 6} )}{8.85 \cdot 1 0 ^{- 12}}

What are the solutions to the equation 19j^2+31j=0 ?
The solutions to the equation 19j^2+31j=0 are j=0,j=-31/19
Find the zeros of 19j^2+31j=0
The zeros of 19j^2+31j=0 are j=0,j=-31/19