What are the solutions to the equation 2x^2-x+10=0 ?

The solutions to the equation 2x^2-x+10=0 are x= 1/4+i(sqrt(79))/4 ,x= 1/4-i(sqrt(79))/4

Find the zeros of 2x^2-x+10=0

The zeros of 2x^2-x+10=0 are x= 1/4+i(sqrt(79))/4 ,x= 1/4-i(sqrt(79))/4

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2 x^{2} - x + 10 = 0

x = \frac{1}{4} + i \frac{79}{4}, x = \frac{1}{4} - i \frac{79}{4}

Solution steps

2 x^{2} - x + 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 2 \cdot 10}{2 \cdot 2}

Simplify

(- 1)^{2} - 4 \cdot 2 \cdot 10 : 79 i

x_{1, 2} = \frac{- ( - 1 ) \pm 79 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 79 i}{2 \cdot 2}, x_{2} = \frac{- ( - 1 ) - 79 i}{2 \cdot 2}

x = \frac{- ( - 1 ) + 79 i}{2 \cdot 2} : \frac{1}{4} + i \frac{79}{4}

x = \frac{- ( - 1 ) - 79 i}{2 \cdot 2} : \frac{1}{4} - i \frac{79}{4}

The solutions to the quadratic equation are:

x = \frac{1}{4} + i \frac{79}{4}, x = \frac{1}{4} - i \frac{79}{4}

96 = 2 x + (2 x + 12)

2^{\frac{4}{2}}

∣ 5 x ∣ \leq 10

6 y + 2 = 9 y + 14

2 < - 2 - p

What are the solutions to the equation 2x^2-x+10=0 ?
The solutions to the equation 2x^2-x+10=0 are x= 1/4+i(sqrt(79))/4 ,x= 1/4-i(sqrt(79))/4
Find the zeros of 2x^2-x+10=0
The zeros of 2x^2-x+10=0 are x= 1/4+i(sqrt(79))/4 ,x= 1/4-i(sqrt(79))/4