What are the solutions to the equation 8x^2+x-3=0 ?

The solutions to the equation 8x^2+x-3=0 are x=(-1+sqrt(97))/(16),x=(-1-sqrt(97))/(16)

Find the zeros of 8x^2+x-3=0

The zeros of 8x^2+x-3=0 are x=(-1+sqrt(97))/(16),x=(-1-sqrt(97))/(16)

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8 x^{2} + x - 3 = 0

x = \frac{- 1 + 97}{16}, x = \frac{- 1 - 97}{16}

Decimal Notation

x = 0.55305 \dots, x = - 0.67805 \dots

Solution steps

8 x^{2} + x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 8 ( - 3 )}{2 \cdot 8}

1^{2} - 4 \cdot 8 (- 3) = 97

x_{1, 2} = \frac{- 1 \pm 97}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- 1 + 97}{2 \cdot 8}, x_{2} = \frac{- 1 - 97}{2 \cdot 8}

x = \frac{- 1 + 97}{2 \cdot 8} : \frac{- 1 + 97}{16}

x = \frac{- 1 - 97}{2 \cdot 8} : \frac{- 1 - 97}{16}

The solutions to the quadratic equation are:

x = \frac{- 1 + 97}{16}, x = \frac{- 1 - 97}{16}

z^{2} - 10 z + 24 = 0

(x - 1)^{2} = 2 x + 6

x^{2} + 2 x - 3

12 m^{2} - 5 m - 2 = 0

16 (1 - 2 x)^{2} - 6 = - 5

What are the solutions to the equation 8x^2+x-3=0 ?
The solutions to the equation 8x^2+x-3=0 are x=(-1+sqrt(97))/(16),x=(-1-sqrt(97))/(16)
Find the zeros of 8x^2+x-3=0
The zeros of 8x^2+x-3=0 are x=(-1+sqrt(97))/(16),x=(-1-sqrt(97))/(16)