What are the solutions to the equation 40x^2+10x-500=0 ?

The solutions to the equation 40x^2+10x-500=0 are x=(-1+3sqrt(89))/8 ,x=-(1+3sqrt(89))/8

Find the zeros of 40x^2+10x-500=0

The zeros of 40x^2+10x-500=0 are x=(-1+3sqrt(89))/8 ,x=-(1+3sqrt(89))/8

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40 x^{2} + 10 x - 500 = 0

x = \frac{- 1 + 3 89}{8}, x = - \frac{1 + 3 89}{8}

Decimal Notation

x = 3.41274 \dots, x = - 3.66274 \dots

Solution steps

40 x^{2} + 10 x - 500 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 40 ( - 500 )}{2 \cdot 40}

1 0^{2} - 4 \cdot 40 (- 500) = 3089

x_{1, 2} = \frac{- 10 \pm 30 89}{2 \cdot 40}

Separate the solutions

x_{1} = \frac{- 10 + 30 89}{2 \cdot 40}, x_{2} = \frac{- 10 - 30 89}{2 \cdot 40}

x = \frac{- 10 + 30 89}{2 \cdot 40} : \frac{- 1 + 3 89}{8}

x = \frac{- 10 - 30 89}{2 \cdot 40} : - \frac{1 + 3 89}{8}

The solutions to the quadratic equation are:

x = \frac{- 1 + 3 89}{8}, x = - \frac{1 + 3 89}{8}

x^{2} - 2 x + 1 = 4

\frac{3}{11}

- x^{4} + 2 x^{3} - 2 x + 1

2 (2) + (8) = - 4

\frac{6 c ^{2} + 13 c - 63}{6 c ^{2} - 17 c + 7} \cdot \frac{2 c ^{2} - 9 c + 4}{12 c + 54}

What are the solutions to the equation 40x^2+10x-500=0 ?
The solutions to the equation 40x^2+10x-500=0 are x=(-1+3sqrt(89))/8 ,x=-(1+3sqrt(89))/8
Find the zeros of 40x^2+10x-500=0
The zeros of 40x^2+10x-500=0 are x=(-1+3sqrt(89))/8 ,x=-(1+3sqrt(89))/8