What are the solutions to the equation 14x^2+19x-3=0 ?

The solutions to the equation 14x^2+19x-3=0 are x= 1/7 ,x=-3/2

Find the zeros of 14x^2+19x-3=0

The zeros of 14x^2+19x-3=0 are x= 1/7 ,x=-3/2

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14 x^{2} + 19 x - 3 = 0

x = \frac{1}{7}, x = - \frac{3}{2}

Decimal Notation

x = 0.14285 \dots, x = - 1.5

Solution steps

14 x^{2} + 19 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 19 \pm 1 9 ^{2} - 4 \cdot 14 ( - 3 )}{2 \cdot 14}

1 9^{2} - 4 \cdot 14 (- 3) = 23

x_{1, 2} = \frac{- 19 \pm 23}{2 \cdot 14}

Separate the solutions

x_{1} = \frac{- 19 + 23}{2 \cdot 14}, x_{2} = \frac{- 19 - 23}{2 \cdot 14}

x = \frac{- 19 + 23}{2 \cdot 14} : \frac{1}{7}

x = \frac{- 19 - 23}{2 \cdot 14} : - \frac{3}{2}

The solutions to the quadratic equation are:

x = \frac{1}{7}, x = - \frac{3}{2}

\frac{9}{7 + 3}

x^{2} - 56 = x

4 x^{2} + 36 x + 80

\frac{5 ^{3}}{5 ^{5}}

84 - 8 f = 580

What are the solutions to the equation 14x^2+19x-3=0 ?
The solutions to the equation 14x^2+19x-3=0 are x= 1/7 ,x=-3/2
Find the zeros of 14x^2+19x-3=0
The zeros of 14x^2+19x-3=0 are x= 1/7 ,x=-3/2