Solve for x
x=\frac{\left(y+3\right)^{2}-7}{2}
y+3\geq 0
Solve for x (complex solution)
x=\frac{\left(y+3\right)^{2}-7}{2}
y=-3\text{ or }arg(y+3)<\pi
Solve for y
y=\sqrt{2x+7}-3
x\geq -\frac{7}{2}
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\sqrt{2x+7}-3=y
Swap sides so that all variable terms are on the left hand side.
\sqrt{2x+7}=y+3
Add 3 to both sides.
2x+7=\left(y+3\right)^{2}
Square both sides of the equation.
2x+7-7=\left(y+3\right)^{2}-7
Subtract 7 from both sides of the equation.
2x=\left(y+3\right)^{2}-7
Subtracting 7 from itself leaves 0.
\frac{2x}{2}=\frac{\left(y+3\right)^{2}-7}{2}
Divide both sides by 2.
x=\frac{\left(y+3\right)^{2}-7}{2}
Dividing by 2 undoes the multiplication by 2.
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