What is the answer to solvefor d,x=sqrt(d)+2 ?

The answer to solvefor d,x=sqrt(d)+2 is d=x^2-4x+4 {x>= 2}

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solvefor d,x=sqrt(d)+2

solve for

d, x = d + 2

d = x^{2} - 4 x + 4 {x \geq 2}

Solution steps

x = d + 2

Switch sides

d + 2 = x

Move

2

to the right side

d = x - 2

Square both sides:

d = x^{2} - 4 x + 4

d = x^{2} - 4 x + 4

Verify Solutions:

d = x^{2} - 4 x + 4 {x \geq 2}

The solution is

d = x^{2} - 4 x + 4 {x \geq 2}

x + 4 + x = 4

3 x = 15 + 3 x + 1

2 x + 7 = 11

4 x - 9 + 1 = 2 x

3 x + 43 = x + 1

What is the answer to solvefor d,x=sqrt(d)+2 ?
The answer to solvefor d,x=sqrt(d)+2 is d=x^2-4x+4 {x>= 2}