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

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

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

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

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solve for

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

y = \frac{3 + 4 x + 1}{2}, y = \frac{3 - 4 x + 1}{2}

Solution steps

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

Switch sides

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

Move

x

to the left side

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

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

y = \frac{- ( - 3 ) + 4 x + 1}{2 \cdot 1} : \frac{3 + 4 x + 1}{2}

y = \frac{- ( - 3 ) - 4 x + 1}{2 \cdot 1} : \frac{3 - 4 x + 1}{2}

The solutions to the quadratic equation are:

y = \frac{3 + 4 x + 1}{2}, y = \frac{3 - 4 x + 1}{2}

\frac{x ^{2}}{4} = 25

x^{2} = (2 x - 9)^{2}

s^{2} - 1 = 3

3 x^{2} - 32 x + 77 = 0

\frac{3}{4} x^{2} - 12 = 0

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