What are the solutions to the equation 8x^2-5x+2=0 ?

The solutions to the equation 8x^2-5x+2=0 are x= 5/16+i(sqrt(39))/(16),x= 5/16-i(sqrt(39))/(16)

Find the zeros of 8x^2-5x+2=0

The zeros of 8x^2-5x+2=0 are x= 5/16+i(sqrt(39))/(16),x= 5/16-i(sqrt(39))/(16)

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8 x^{2} - 5 x + 2 = 0

x = \frac{5}{16} + i \frac{39}{16}, x = \frac{5}{16} - i \frac{39}{16}

Solution steps

8 x^{2} - 5 x + 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 8 \cdot 2}{2 \cdot 8}

Simplify

(- 5)^{2} - 4 \cdot 8 \cdot 2 : 39 i

x_{1, 2} = \frac{- ( - 5 ) \pm 39 i}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 39 i}{2 \cdot 8}, x_{2} = \frac{- ( - 5 ) - 39 i}{2 \cdot 8}

x = \frac{- ( - 5 ) + 39 i}{2 \cdot 8} : \frac{5}{16} + i \frac{39}{16}

x = \frac{- ( - 5 ) - 39 i}{2 \cdot 8} : \frac{5}{16} - i \frac{39}{16}

The solutions to the quadratic equation are:

x = \frac{5}{16} + i \frac{39}{16}, x = \frac{5}{16} - i \frac{39}{16}

4.5 x + 27 = - 9

- \frac{5}{4} + \frac{1}{6}

f (n) = \frac{n ^{3} - n}{n ^{2} - 5 n - 6}

\frac{5}{3} + (- \frac{7}{6})

\frac{3}{5} x + \frac{4}{5} = - \frac{2}{5}

What are the solutions to the equation 8x^2-5x+2=0 ?
The solutions to the equation 8x^2-5x+2=0 are x= 5/16+i(sqrt(39))/(16),x= 5/16-i(sqrt(39))/(16)
Find the zeros of 8x^2-5x+2=0
The zeros of 8x^2-5x+2=0 are x= 5/16+i(sqrt(39))/(16),x= 5/16-i(sqrt(39))/(16)