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

The solutions to the equation 2x^2-x=-5 are x= 1/4+i(sqrt(39))/4 ,x= 1/4-i(sqrt(39))/4

Find the zeros of 2x^2-x=-5

The zeros of 2x^2-x=-5 are x= 1/4+i(sqrt(39))/4 ,x= 1/4-i(sqrt(39))/4

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2 x^{2} - x = - 5

x = \frac{1}{4} + i \frac{39}{4}, x = \frac{1}{4} - i \frac{39}{4}

Solution steps

2 x^{2} - x = - 5

Move

5

to the left side

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

Solve with the quadratic formula

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

Simplify

(- 1)^{2} - 4 \cdot 2 \cdot 5 : 39 i

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

Separate the solutions

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

x = \frac{- ( - 1 ) + 39 i}{2 \cdot 2} : \frac{1}{4} + i \frac{39}{4}

x = \frac{- ( - 1 ) - 39 i}{2 \cdot 2} : \frac{1}{4} - i \frac{39}{4}

The solutions to the quadratic equation are:

x = \frac{1}{4} + i \frac{39}{4}, x = \frac{1}{4} - i \frac{39}{4}

x, (x^{2}) + (x^{2}) = (y^{2})

2 x^{2} - 12 x + 19

3 x^{2} + 7 x = 0

- 14 x^{2} + 3 x - 2 = 0

x^{2} + 9

What are the solutions to the equation 2x^2-x=-5 ?
The solutions to the equation 2x^2-x=-5 are x= 1/4+i(sqrt(39))/4 ,x= 1/4-i(sqrt(39))/4
Find the zeros of 2x^2-x=-5
The zeros of 2x^2-x=-5 are x= 1/4+i(sqrt(39))/4 ,x= 1/4-i(sqrt(39))/4