What are the solutions to the equation 3y-y^2-4=0 ?

The solutions to the equation 3y-y^2-4=0 are y= 3/2-i(sqrt(7))/2 ,y= 3/2+i(sqrt(7))/2

Find the zeros of 3y-y^2-4=0

The zeros of 3y-y^2-4=0 are y= 3/2-i(sqrt(7))/2 ,y= 3/2+i(sqrt(7))/2

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3 y - y^{2} - 4 = 0

y = \frac{3}{2} - i \frac{7}{2}, y = \frac{3}{2} + i \frac{7}{2}

Solution steps

3 y - y^{2} - 4 = 0

Write in the standard form

a x^{2} + b x + c = 0

- y^{2} + 3 y - 4 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 1 ) ( - 4 )}{2 ( - 1 )}

Simplify

3^{2} - 4 (- 1) (- 4) : 7 i

y_{1, 2} = \frac{- 3 \pm 7 i}{2 ( - 1 )}

Separate the solutions

y_{1} = \frac{- 3 + 7 i}{2 ( - 1 )}, y_{2} = \frac{- 3 - 7 i}{2 ( - 1 )}

y = \frac{- 3 + 7 i}{2 ( - 1 )} : \frac{3}{2} - i \frac{7}{2}

y = \frac{- 3 - 7 i}{2 ( - 1 )} : \frac{3}{2} + i \frac{7}{2}

The solutions to the quadratic equation are:

y = \frac{3}{2} - i \frac{7}{2}, y = \frac{3}{2} + i \frac{7}{2}

s^{2} + 14 s + 2 = 0

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

- x - x^{2} = 0

x^{2} + 4 x = 12

6 x^{2} - 35 = 11 x

What are the solutions to the equation 3y-y^2-4=0 ?
The solutions to the equation 3y-y^2-4=0 are y= 3/2-i(sqrt(7))/2 ,y= 3/2+i(sqrt(7))/2
Find the zeros of 3y-y^2-4=0
The zeros of 3y-y^2-4=0 are y= 3/2-i(sqrt(7))/2 ,y= 3/2+i(sqrt(7))/2