What are the solutions to the equation solvefor y,x=y^2-4y+3 ?

The solutions to the equation solvefor y,x=y^2-4y+3 are y=2+sqrt(x+1),y=2-sqrt(x+1)

Find the zeros of solvefor y,x=y^2-4y+3

The zeros of solvefor y,x=y^2-4y+3 are y=2+sqrt(x+1),y=2-sqrt(x+1)

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solve for

y, x = y^{2} - 4 y + 3

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

Solution steps

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

Switch sides

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

Move

x

to the left side

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

Solve with the quadratic formula

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

Simplify

(- 4)^{2} - 4 \cdot 1 \cdot (3 - 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

(8 x + 9)^{2} = 4

x, x^{2} + y^{2} + 6 = 0

- x^{2} - 12 = - 8 x

43575 - 75 xx = 4275

3 x^{2} - 2 x - 14 = 2

What are the solutions to the equation solvefor y,x=y^2-4y+3 ?
The solutions to the equation solvefor y,x=y^2-4y+3 are y=2+sqrt(x+1),y=2-sqrt(x+1)
Find the zeros of solvefor y,x=y^2-4y+3
The zeros of solvefor y,x=y^2-4y+3 are y=2+sqrt(x+1),y=2-sqrt(x+1)