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

The solutions to the equation y^2-y+2=0 are y= 1/2+i(sqrt(7))/2 ,y= 1/2-i(sqrt(7))/2

Find the zeros of y^2-y+2=0

The zeros of y^2-y+2=0 are y= 1/2+i(sqrt(7))/2 ,y= 1/2-i(sqrt(7))/2

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y^{2} - y + 2 = 0

y = \frac{1}{2} + i \frac{7}{2}, y = \frac{1}{2} - i \frac{7}{2}

Solution steps

y^{2} - y + 2 = 0

Solve with the quadratic formula

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

Simplify

(- 1)^{2} - 4 \cdot 1 \cdot 2 : 7 i

y_{1, 2} = \frac{- ( - 1 ) \pm 7 i}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 1 ) + 7 i}{2 \cdot 1}, y_{2} = \frac{- ( - 1 ) - 7 i}{2 \cdot 1}

y = \frac{- ( - 1 ) + 7 i}{2 \cdot 1} : \frac{1}{2} + i \frac{7}{2}

y = \frac{- ( - 1 ) - 7 i}{2 \cdot 1} : \frac{1}{2} - i \frac{7}{2}

The solutions to the quadratic equation are:

y = \frac{1}{2} + i \frac{7}{2}, y = \frac{1}{2} - i \frac{7}{2}

p^{2} - 13 p - 14 = 0

5 (m - 3) + 2 m = 27

\frac{2 x + 3}{4} = \frac{2 x - 5}{6} + 1

\frac{3 ^{3} \cdot 3 ^{2}}{3 ^{3}}

\frac{a ^{5}}{a ^{2}}

What are the solutions to the equation y^2-y+2=0 ?
The solutions to the equation y^2-y+2=0 are y= 1/2+i(sqrt(7))/2 ,y= 1/2-i(sqrt(7))/2
Find the zeros of y^2-y+2=0
The zeros of y^2-y+2=0 are y= 1/2+i(sqrt(7))/2 ,y= 1/2-i(sqrt(7))/2