What are the solutions to the equation 4x^2+x-6=0 ?

The solutions to the equation 4x^2+x-6=0 are x=(-1+sqrt(97))/8 ,x=(-1-sqrt(97))/8

Find the zeros of 4x^2+x-6=0

The zeros of 4x^2+x-6=0 are x=(-1+sqrt(97))/8 ,x=(-1-sqrt(97))/8

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4 x^{2} + x - 6 = 0

x = \frac{- 1 + 97}{8}, x = \frac{- 1 - 97}{8}

Decimal Notation

x = 1.10610 \dots, x = - 1.35610 \dots

Solution steps

4 x^{2} + x - 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 4 ( - 6 )}{2 \cdot 4}

1^{2} - 4 \cdot 4 (- 6) = 97

x_{1, 2} = \frac{- 1 \pm 97}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 1 + 97}{2 \cdot 4}, x_{2} = \frac{- 1 - 97}{2 \cdot 4}

x = \frac{- 1 + 97}{2 \cdot 4} : \frac{- 1 + 97}{8}

x = \frac{- 1 - 97}{2 \cdot 4} : \frac{- 1 - 97}{8}

The solutions to the quadratic equation are:

x = \frac{- 1 + 97}{8}, x = \frac{- 1 - 97}{8}

1 = - 88 (\frac{3 - 4 b}{56}) - 4 b

i^{55}

9 - \frac{1}{2}

- 4 < \frac{q - 2}{- 2}

2147

What are the solutions to the equation 4x^2+x-6=0 ?
The solutions to the equation 4x^2+x-6=0 are x=(-1+sqrt(97))/8 ,x=(-1-sqrt(97))/8
Find the zeros of 4x^2+x-6=0
The zeros of 4x^2+x-6=0 are x=(-1+sqrt(97))/8 ,x=(-1-sqrt(97))/8