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

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

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

The zeros of 2x^2-x+1/8 =0 are x= 1/4

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

x = \frac{1}{4}

Decimal Notation

x = 0.25

Solution steps

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

Multiply both sides by

8

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 16 \cdot 1}{2 \cdot 16}

(- 8)^{2} - 4 \cdot 16 \cdot 1 = 0

x_{1, 2} = \frac{- ( - 8 ) \pm 0}{2 \cdot 16}

x = \frac{- ( - 8 )}{2 \cdot 16}

\frac{- ( - 8 )}{2 \cdot 16} = \frac{1}{4}

x = \frac{1}{4}

The solution to the quadratic equation is:

x = \frac{1}{4}

x = 6 \cdot 1 0^{- 3} (1 - \frac{( 1600 \cdot x ) + 3}{6})^{2}

0.2 q^{2} + 4 q - 60 = 0

0.35 x^{2} - x + 0.35 = 0

6 x^{2} - 12 x = 720

1.034 s^{2} + 2.67 s + 2 = 0

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