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

The solutions to the equation 2x^2-x=-2 are x= 1/4+i(sqrt(15))/4 ,x= 1/4-i(sqrt(15))/4

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

The zeros of 2x^2-x=-2 are x= 1/4+i(sqrt(15))/4 ,x= 1/4-i(sqrt(15))/4

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2 x^{2} - x = - 2

x = \frac{1}{4} + i \frac{15}{4}, x = \frac{1}{4} - i \frac{15}{4}

Solution steps

2 x^{2} - x = - 2

Move

2

to the left side

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

Solve with the quadratic formula

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

Simplify

(- 1)^{2} - 4 \cdot 2 \cdot 2 : 15 i

x_{1, 2} = \frac{- ( - 1 ) \pm 15 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 15 i}{2 \cdot 2}, x_{2} = \frac{- ( - 1 ) - 15 i}{2 \cdot 2}

x = \frac{- ( - 1 ) + 15 i}{2 \cdot 2} : \frac{1}{4} + i \frac{15}{4}

x = \frac{- ( - 1 ) - 15 i}{2 \cdot 2} : \frac{1}{4} - i \frac{15}{4}

The solutions to the quadratic equation are:

x = \frac{1}{4} + i \frac{15}{4}, x = \frac{1}{4} - i \frac{15}{4}

\frac{x}{4} - \frac{1}{4} \leq \frac{x}{5} + 1

- \frac{1}{2} + 5

7 i^{4} - 12 i^{2}

\frac{2}{3} x + 2 < \frac{1}{6} x - 1

lo g_{3} (6) + lo g_{3} (4.5)

What are the solutions to the equation 2x^2-x=-2 ?
The solutions to the equation 2x^2-x=-2 are x= 1/4+i(sqrt(15))/4 ,x= 1/4-i(sqrt(15))/4
Find the zeros of 2x^2-x=-2
The zeros of 2x^2-x=-2 are x= 1/4+i(sqrt(15))/4 ,x= 1/4-i(sqrt(15))/4