(1/2)b=3

Solution for (1/2)b=3 equation:

(1/2)b=3

We move all terms to the left:

(1/2)b-(3)=0


Domain of the equation: 2)b!=0

b!=0/1

b!=0

b∈R


We add all the numbers together, and all the variables

(+1/2)b-3=0

We multiply parentheses

b^2-3=0

a = 1; b = 0; c = -3;
Δ = b2-4ac
Δ = 02-4·1·(-3)
Δ = 12
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{3}}{2*1}=\frac{0-2\sqrt{3}}{2}
=-\frac{2\sqrt{3}}{2}
=-\sqrt{3}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{3}}{2*1}=\frac{0+2\sqrt{3}}{2}
=\frac{2\sqrt{3}}{2}
=\sqrt{3}
$