Solve (4+3^2+6^0)*x=sqrt{100}+2^2

$$( 4 + 3 ^ { 2 } + 6 ^ { 0 } ) \cdot x = \sqrt { 100 } + 2 ^ { 2 }$$

Solve for x

$$x=1$$

Steps for Solving Linear Equation

Calculate 3 to the power of 2 and get 9.

$$\left(4+9+6^{0}\right)x=\sqrt{100}+2^{2}$$

Add 4 and 9 to get 13.

$$\left(13+6^{0}\right)x=\sqrt{100}+2^{2}$$

Calculate 6 to the power of 0 and get 1.

$$\left(13+1\right)x=\sqrt{100}+2^{2}$$

Add 13 and 1 to get 14.

$$14x=\sqrt{100}+2^{2}$$

Calculate the square root of 100 and get 10.

$$14x=10+2^{2}$$

Calculate 2 to the power of 2 and get 4.

$$14x=10+4$$

Add 10 and 4 to get 14.

$$14x=14$$

Divide both sides by 14.

$$x=\frac{14}{14}$$

Divide 14 by 14 to get 1.

$$x=1$$