What are the solutions to the equation -(y+2)(y-1)=3 ?

The solutions to the equation -(y+2)(y-1)=3 are y=-1/2-i(sqrt(3))/2 ,y=-1/2+i(sqrt(3))/2

Find the zeros of -(y+2)(y-1)=3

The zeros of -(y+2)(y-1)=3 are y=-1/2-i(sqrt(3))/2 ,y=-1/2+i(sqrt(3))/2

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- (y + 2) (y - 1) = 3

y = - \frac{1}{2} - i \frac{3}{2}, y = - \frac{1}{2} + i \frac{3}{2}

Solution steps

- (y + 2) (y - 1) = 3

Expand

- (y + 2) (y - 1) : - y^{2} - y + 2

- y^{2} - y + 2 = 3

Move

3

to the left side

- y^{2} - y - 1 = 0

Solve with the quadratic formula

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

Simplify

(- 1)^{2} - 4 (- 1) (- 1) : 3 i

y_{1, 2} = \frac{- ( - 1 ) \pm 3 i}{2 ( - 1 )}

Separate the solutions

y_{1} = \frac{- ( - 1 ) + 3 i}{2 ( - 1 )}, y_{2} = \frac{- ( - 1 ) - 3 i}{2 ( - 1 )}

y = \frac{- ( - 1 ) + 3 i}{2 ( - 1 )} : - \frac{1}{2} - i \frac{3}{2}

y = \frac{- ( - 1 ) - 3 i}{2 ( - 1 )} : - \frac{1}{2} + i \frac{3}{2}

The solutions to the quadratic equation are:

y = - \frac{1}{2} - i \frac{3}{2}, y = - \frac{1}{2} + i \frac{3}{2}

5 x^{2} + 8 = - 27

100 x^{2} + 200 x + 100 = 0

x^{2} + 14 = - 9 x

36 n^{2} - 13 = 12

144 = (n + 1)^{2}

What are the solutions to the equation -(y+2)(y-1)=3 ?
The solutions to the equation -(y+2)(y-1)=3 are y=-1/2-i(sqrt(3))/2 ,y=-1/2+i(sqrt(3))/2
Find the zeros of -(y+2)(y-1)=3
The zeros of -(y+2)(y-1)=3 are y=-1/2-i(sqrt(3))/2 ,y=-1/2+i(sqrt(3))/2