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 = 0.67805 \dots, x = - 0.55305 \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}

\frac{4}{3} \cdot 3

3 (4 - 2 x) = x - 2

(\frac{1}{4})^{\frac{3}{2}}

(x^{2} - 3 x + 2) (x^{2} + 7 x - 8) \leq 0

\frac{s}{2} - 1 \leq 1

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)