What is the answer to solvefor x,sqrt(1+x+1)=y ?

The answer to solvefor x,sqrt(1+x+1)=y is x=y^2-2 {y>= 0}

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solvefor x,sqrt(1+x+1)=y

solve for

x, 1 + x + 1 = y

x = y^{2} - 2 {y \geq 0}

Solution steps

1 + x + 1 = y

Square both sides:

x + 2 = y^{2}

x + 2 = y^{2}

Solve

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

x = y^{2} - 2

Verify Solutions:

x = y^{2} - 2 {y \geq 0}

The solution is

x = y^{2} - 2 {y \geq 0}

3 y = 2

5 x - 6 = - 1

\frac{x}{4} = 3

x \frac{a ( 1 + x )}{x - 1} = b

(x - 5)^{\frac{4}{3}} = 16

What is the answer to solvefor x,sqrt(1+x+1)=y ?
The answer to solvefor x,sqrt(1+x+1)=y is x=y^2-2 {y>= 0}