Solve A=sqrt{(a^2+6a+a)-(a+4)(a+2)}

$$A = \sqrt { ( a ^ { 2 } + 6 a + a ) - ( a + 4 ) ( a + 2 ) }$$

Solve for a

$$a=A^{2}+8$$

$$A\geq 0$$

Steps for Solving Radical Equations

Combine 6a and a to get 7a.

$$A=\sqrt{a^{2}+7a-\left(a+4\right)\left(a+2\right)}$$

Use the distributive property to multiply a+4 by a+2 and combine like terms.

$$A=\sqrt{a^{2}+7a-\left(a^{2}+6a+8\right)}$$

To find the opposite of a^{2}+6a+8, find the opposite of each term.

$$A=\sqrt{a^{2}+7a-a^{2}-6a-8}$$

Combine a^{2} and -a^{2} to get 0.

$$A=\sqrt{7a-6a-8}$$

Combine 7a and -6a to get a.

$$A=\sqrt{a-8}$$

Swap sides so that all variable terms are on the left hand side.

$$\sqrt{a-8}=A$$

Square both sides of the equation.

$$a-8=A^{2}$$

Add 8 to both sides of the equation.

$$a-8-\left(-8\right)=A^{2}-\left(-8\right)$$

Subtracting -8 from itself leaves 0.

$$a=A^{2}-\left(-8\right)$$

Subtract -8 from A^{2}.

$$a=A^{2}+8$$

Solve for A

$$A=\sqrt{a-8}$$

$$a\geq 8$$