Solve sqrt{a+b+c}+sqrt{(a^2+1)(b-b)}+|10-2b|=2

$$\sqrt { a + b + c } + \sqrt { ( a ^ { 2 } + 1 ) ( b - b ) } + | 10 - 2 b | = 2$$

Solve for a

$$\left\{\begin{matrix}a=-\left(-4\left(4-b\right)^{2}+b+c\right)\text{, }&-\left(8-2b\right)\geq 0\text{ and }b\leq 5\\a=-\left(-4\left(6-b\right)^{2}+b+c\right)\text{, }&12-2b\geq 0\text{ and }b\geq 5\end{matrix}\right.$$

Steps for Solving Radical Equations

Combine b and -b to get 0.

$$\sqrt{a+b+c}+\sqrt{\left(a^{2}+1\right)\times 0}+|10-2b|=2$$

Anything times zero gives zero.

$$\sqrt{a+b+c}+\sqrt{0}+|10-2b|=2$$

Calculate the square root of 0 and get 0.

$$\sqrt{a+b+c}+0+|10-2b|=2$$

Anything plus zero gives itself.

$$\sqrt{a+b+c}+|10-2b|=2$$

Subtract |10-2b| from both sides.

$$\sqrt{a+b+c}=2-|10-2b|$$

Square both sides of the equation.

$$a+b+c=\left(-|10-2b|+2\right)^{2}$$

Subtract b+c from both sides of the equation.

$$a+b+c-\left(b+c\right)=\left(-|10-2b|+2\right)^{2}-\left(b+c\right)$$

Subtracting b+c from itself leaves 0.

$$a=\left(-|10-2b|+2\right)^{2}-\left(b+c\right)$$

Subtract b+c from \left(2-|10-2b|\right)^{2}.

$$a=\left(-|10-2b|+2\right)^{2}-b-c$$

Solve for b

$$\left\{\begin{matrix}b=\frac{\sqrt{16a+16c+65}+33}{8}\text{, }&a\geq -c-\frac{65}{16}\text{ and }a\leq -c-1\\b=\frac{-\sqrt{16a+16c+65}+33}{8}\text{, }&a\geq -c-\frac{65}{16}\text{ and }a\leq -c-4\\b=\frac{-\sqrt{16a+16c+97}+49}{8}\text{, }&a\geq -c-6\text{ and }a\leq -c-1\end{matrix}\right.$$