What are the solutions to the equation solvefor x,x^2-5x+6=y ?

The solutions to the equation solvefor x,x^2-5x+6=y are x=(5+sqrt(4y+1))/2 ,x=(5-sqrt(4y+1))/2

Find the zeros of solvefor x,x^2-5x+6=y

The zeros of solvefor x,x^2-5x+6=y are x=(5+sqrt(4y+1))/2 ,x=(5-sqrt(4y+1))/2

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solve for

x, x^{2} - 5 x + 6 = y

x = \frac{5 + 4 y + 1}{2}, x = \frac{5 - 4 y + 1}{2}

Solution steps

x^{2} - 5 x + 6 = y

Move

y

to the left side

x^{2} - 5 x + 6 - y = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot ( 6 - y )}{2 \cdot 1}

Simplify

(- 5)^{2} - 4 \cdot 1 \cdot (6 - y) : 4 y + 1

x_{1, 2} = \frac{- ( - 5 ) \pm 4 y + 1}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 4 y + 1}{2 \cdot 1}, x_{2} = \frac{- ( - 5 ) - 4 y + 1}{2 \cdot 1}

x = \frac{- ( - 5 ) + 4 y + 1}{2 \cdot 1} : \frac{5 + 4 y + 1}{2}

x = \frac{- ( - 5 ) - 4 y + 1}{2 \cdot 1} : \frac{5 - 4 y + 1}{2}

The solutions to the quadratic equation are:

x = \frac{5 + 4 y + 1}{2}, x = \frac{5 - 4 y + 1}{2}

3 w (w - 2) = 10

x^{2} - 17 x - 18 = 0

3 x^{2} = - 8 x + 8

9 x^{2} - 5 x = 1

- 8 x^{2} - 4 x + 24 = 0

What are the solutions to the equation solvefor x,x^2-5x+6=y ?
The solutions to the equation solvefor x,x^2-5x+6=y are x=(5+sqrt(4y+1))/2 ,x=(5-sqrt(4y+1))/2
Find the zeros of solvefor x,x^2-5x+6=y
The zeros of solvefor x,x^2-5x+6=y are x=(5+sqrt(4y+1))/2 ,x=(5-sqrt(4y+1))/2