What are the solutions to the equation 2x^2-5x+67=0 ?

The solutions to the equation 2x^2-5x+67=0 are x= 5/4+i(sqrt(511))/4 ,x= 5/4-i(sqrt(511))/4

Find the zeros of 2x^2-5x+67=0

The zeros of 2x^2-5x+67=0 are x= 5/4+i(sqrt(511))/4 ,x= 5/4-i(sqrt(511))/4

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2 x^{2} - 5 x + 67 = 0

x = \frac{5}{4} + i \frac{511}{4}, x = \frac{5}{4} - i \frac{511}{4}

Solution steps

2 x^{2} - 5 x + 67 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 2 \cdot 67}{2 \cdot 2}

Simplify

(- 5)^{2} - 4 \cdot 2 \cdot 67 : 511 i

x_{1, 2} = \frac{- ( - 5 ) \pm 511 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 511 i}{2 \cdot 2}, x_{2} = \frac{- ( - 5 ) - 511 i}{2 \cdot 2}

x = \frac{- ( - 5 ) + 511 i}{2 \cdot 2} : \frac{5}{4} + i \frac{511}{4}

x = \frac{- ( - 5 ) - 511 i}{2 \cdot 2} : \frac{5}{4} - i \frac{511}{4}

The solutions to the quadratic equation are:

x = \frac{5}{4} + i \frac{511}{4}, x = \frac{5}{4} - i \frac{511}{4}

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What are the solutions to the equation 2x^2-5x+67=0 ?
The solutions to the equation 2x^2-5x+67=0 are x= 5/4+i(sqrt(511))/4 ,x= 5/4-i(sqrt(511))/4
Find the zeros of 2x^2-5x+67=0
The zeros of 2x^2-5x+67=0 are x= 5/4+i(sqrt(511))/4 ,x= 5/4-i(sqrt(511))/4