What are the solutions to the equation 4x^2+x-1/8 =0 ?

The solutions to the equation 4x^2+x-1/8 =0 are x=(-1+sqrt(3))/8 ,x=-(1+sqrt(3))/8

Find the zeros of 4x^2+x-1/8 =0

The zeros of 4x^2+x-1/8 =0 are x=(-1+sqrt(3))/8 ,x=-(1+sqrt(3))/8

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4 x^{2} + x - \frac{1}{8} = 0

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

Decimal Notation

x = 0.09150 \dots, x = - 0.34150 \dots

Solution steps

4 x^{2} + x - \frac{1}{8} = 0

Multiply both sides by

8

32 x^{2} + 8 x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 32 ( - 1 )}{2 \cdot 32}

8^{2} - 4 \cdot 32 (- 1) = 83

x_{1, 2} = \frac{- 8 \pm 8 3}{2 \cdot 32}

Separate the solutions

x_{1} = \frac{- 8 + 8 3}{2 \cdot 32}, x_{2} = \frac{- 8 - 8 3}{2 \cdot 32}

x = \frac{- 8 + 8 3}{2 \cdot 32} : \frac{- 1 + 3}{8}

x = \frac{- 8 - 8 3}{2 \cdot 32} : - \frac{1 + 3}{8}

The solutions to the quadratic equation are:

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

r + 5 \leq 11

4 x^{2} - 8 = 32

\frac{25 x y ^{2}}{125 x ^{5} y ^{7}}

\frac{x}{4} = \frac{x}{3} - 2

- \frac{1}{4} (4) + 4

What are the solutions to the equation 4x^2+x-1/8 =0 ?
The solutions to the equation 4x^2+x-1/8 =0 are x=(-1+sqrt(3))/8 ,x=-(1+sqrt(3))/8
Find the zeros of 4x^2+x-1/8 =0
The zeros of 4x^2+x-1/8 =0 are x=(-1+sqrt(3))/8 ,x=-(1+sqrt(3))/8