What are the solutions to the equation solvefor x,x^2-2x-4y+5=0 ?

The solutions to the equation solvefor x,x^2-2x-4y+5=0 are x=1+2sqrt(y-1),x=1-2sqrt(y-1)

Find the zeros of solvefor x,x^2-2x-4y+5=0

The zeros of solvefor x,x^2-2x-4y+5=0 are x=1+2sqrt(y-1),x=1-2sqrt(y-1)

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solve for

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

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

Solution steps

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

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

x^{2} + 10 x + 23

(x - 12)^{2} = 54

y^{2} = - 3

0 = 12 x^{2} - 88 x + 112

2 x^{2} + 7 x + 4 = 8

What are the solutions to the equation solvefor x,x^2-2x-4y+5=0 ?
The solutions to the equation solvefor x,x^2-2x-4y+5=0 are x=1+2sqrt(y-1),x=1-2sqrt(y-1)
Find the zeros of solvefor x,x^2-2x-4y+5=0
The zeros of solvefor x,x^2-2x-4y+5=0 are x=1+2sqrt(y-1),x=1-2sqrt(y-1)