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

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

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

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

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solve for

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

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

Solution steps

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

Move

y

to the left side

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

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 4 y + 1}{2 \cdot 1}, x_{2} = \frac{- ( - 3 ) - 4 y + 1}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

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

x^{2} + 3 = 9

2 x^{2} - 8 x - 5

5 a^{2} - 36 a = 32

x, f = a (x - h)^{2}

- 2 x^{2} + 14 x = 0

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