Solve a=(1/2-sqrt{3})+(1/2+1)+(1-sqrt{5})*(1+sqrt{5})

$$a = ( \frac { 1 } { 2 } - \sqrt { 3 } ) + ( \frac { 1 } { 2 } + 1 ) + ( 1 - \sqrt { 5 } ) \cdot ( 1 + \sqrt { 5 } )$$

Solve for a

$$a=-\left(\sqrt{3}+2\right)\approx -3.732050808$$

Steps for Solving Linear Equation

Since \frac{1}{2} and \frac{1}{2} have the same denominator, add them by adding their numerators.

$$a=\frac{1+1}{2}-\sqrt{3}+1+\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)$$

Add 1 and 1 to get 2.

$$a=\frac{2}{2}-\sqrt{3}+1+\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)$$

Divide 2 by 2 to get 1.

$$a=1-\sqrt{3}+1+\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)$$

Add 1 and 1 to get 2.

$$a=2-\sqrt{3}+\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)$$

Consider \left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$a=2-\sqrt{3}+1^{2}-\left(\sqrt{5}\right)^{2}$$

Calculate 1 to the power of 2 and get 1.

$$a=2-\sqrt{3}+1-\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{5} is 5.

$$a=2-\sqrt{3}+1-5$$

Subtract 5 from 1 to get -4.

$$a=2-\sqrt{3}-4$$

Subtract 4 from 2 to get -2.

$$a=-2-\sqrt{3}$$

Assign a

$$a≔-\sqrt{3}-2$$