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

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

Find the zeros of 2=(y/(2-3))(y-2)

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

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2 = (\frac{y}{2 - 3}) (y - 2)

y = 1 - i, y = 1 + i

Solution steps

2 = (\frac{y}{2 - 3}) (y - 2)

Expand

(\frac{y}{2 - 3}) (y - 2) : - y^{2} + 2 y

2 = - y^{2} + 2 y

Switch sides

- y^{2} + 2 y = 2

Move

2

to the left side

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

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = 1 - i, y = 1 + i

x, a \cdot x^{2} + b \cdot y^{2} + z^{2} = 1

(a - b) x^{2} + (b - c) x + (c - a) = 0

x^{2} - 2 (a^{2} + b^{2}) x + (a^{2} - b^{2}) = 0

x, p^{3} - (x^{2} + x y + y^{2}) \cdot p + x y (x + y) = 0

\frac{5}{2} \cdot x + \frac{2}{3} \cdot x = \frac{95}{2} \cdot x^{2}

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