What are the solutions to the equation solvefor y,x=y^2-4y+5 ?

The solutions to the equation solvefor y,x=y^2-4y+5 are y=2+sqrt(x-1),y=2-sqrt(x-1)

Find the zeros of solvefor y,x=y^2-4y+5

The zeros of solvefor y,x=y^2-4y+5 are y=2+sqrt(x-1),y=2-sqrt(x-1)

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solve for

y, x = y^{2} - 4 y + 5

y = 2 + x - 1, y = 2 - x - 1

Solution steps

x = y^{2} - 4 y + 5

Switch sides

y^{2} - 4 y + 5 = x

Move

x

to the left side

y^{2} - 4 y + 5 - x = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( 5 - x )}{2 \cdot 1}

Simplify

(- 4)^{2} - 4 \cdot 1 \cdot (5 - x) : 2 x - 1

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

Separate the solutions

y_{1} = \frac{- ( - 4 ) + 2 x - 1}{2 \cdot 1}, y_{2} = \frac{- ( - 4 ) - 2 x - 1}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

y = 2 + x - 1, y = 2 - x - 1

x, y = 2 \cdot x - x^{2}

x^{2} - 2 x - 2 = 5

- x^{2} + 4 x + 6

x^{2} - 2 x + 193 = 0

8 (x^{2} + 4) - 2 x - 2 = 2 x^{2} - 5 x + 30

What are the solutions to the equation solvefor y,x=y^2-4y+5 ?
The solutions to the equation solvefor y,x=y^2-4y+5 are y=2+sqrt(x-1),y=2-sqrt(x-1)
Find the zeros of solvefor y,x=y^2-4y+5
The zeros of solvefor y,x=y^2-4y+5 are y=2+sqrt(x-1),y=2-sqrt(x-1)