What is the answer to solvefor y,x=sqrt(y+3)-4 ?

The answer to solvefor y,x=sqrt(y+3)-4 is y=x^2+8x+13 {x>=-4}

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solvefor y,x=sqrt(y+3)-4

solve for

y, x = y + 3 - 4

y = x^{2} + 8 x + 13 {x \geq - 4}

Solution steps

x = y + 3 - 4

Switch sides

y + 3 - 4 = x

Move

4

to the right side

y + 3 = x + 4

Square both sides:

y + 3 = x^{2} + 8 x + 16

y + 3 = x^{2} + 8 x + 16

Solve

y + 3 = x^{2} + 8 x + 16 : y = x^{2} + 8 x + 13

y = x^{2} + 8 x + 13

Verify Solutions:

y = x^{2} + 8 x + 13 {x \geq - 4}

The solution is

y = x^{2} + 8 x + 13 {x \geq - 4}

\frac{1}{x} = 2

3 x - 8 + 1 = x + 5

- x + 5 + 3 x + 4 = 3

3 x + 1 = - 4

4 x - 2 = 2

What is the answer to solvefor y,x=sqrt(y+3)-4 ?
The answer to solvefor y,x=sqrt(y+3)-4 is y=x^2+8x+13 {x>=-4}