Solve for a
a=\left(3-b\right)^{2}+2
-\left(3-b\right)\geq 0\text{ and }b\leq 3
Solve for b
b=3
a=2
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\sqrt{a-2}+|b-3|-|b-3|=-|b-3|
Subtract |b-3| from both sides of the equation.
\sqrt{a-2}=-|b-3|
Subtracting |b-3| from itself leaves 0.
a-2=\left(b-3\right)^{2}
Square both sides of the equation.
a-2-\left(-2\right)=\left(b-3\right)^{2}-\left(-2\right)
Add 2 to both sides of the equation.
a=\left(b-3\right)^{2}-\left(-2\right)
Subtracting -2 from itself leaves 0.
a=\left(b-3\right)^{2}+2
Subtract -2 from \left(b-3\right)^{2}.
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