What are the solutions to the equation 8x^2-3x+1=0 ?

The solutions to the equation 8x^2-3x+1=0 are x= 3/16+i(sqrt(23))/(16),x= 3/16-i(sqrt(23))/(16)

Find the zeros of 8x^2-3x+1=0

The zeros of 8x^2-3x+1=0 are x= 3/16+i(sqrt(23))/(16),x= 3/16-i(sqrt(23))/(16)

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8 x^{2} - 3 x + 1 = 0

x = \frac{3}{16} + i \frac{23}{16}, x = \frac{3}{16} - i \frac{23}{16}

Solution steps

8 x^{2} - 3 x + 1 = 0

Solve with the quadratic formula

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

Simplify

(- 3)^{2} - 4 \cdot 8 \cdot 1 : 23 i

x_{1, 2} = \frac{- ( - 3 ) \pm 23 i}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 23 i}{2 \cdot 8}, x_{2} = \frac{- ( - 3 ) - 23 i}{2 \cdot 8}

x = \frac{- ( - 3 ) + 23 i}{2 \cdot 8} : \frac{3}{16} + i \frac{23}{16}

x = \frac{- ( - 3 ) - 23 i}{2 \cdot 8} : \frac{3}{16} - i \frac{23}{16}

The solutions to the quadratic equation are:

x = \frac{3}{16} + i \frac{23}{16}, x = \frac{3}{16} - i \frac{23}{16}

r, v = n r^{2} h

m^{2} + 36 = 0

x^{2} = 1.5

x^{2} - 5 x - 6 = 0

3 x^{2} - 8 x - 15 = 0

What are the solutions to the equation 8x^2-3x+1=0 ?
The solutions to the equation 8x^2-3x+1=0 are x= 3/16+i(sqrt(23))/(16),x= 3/16-i(sqrt(23))/(16)
Find the zeros of 8x^2-3x+1=0
The zeros of 8x^2-3x+1=0 are x= 3/16+i(sqrt(23))/(16),x= 3/16-i(sqrt(23))/(16)