-(1/2)j=-3

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Solution for -(1/2)j=-3 equation:



-(1/2)j=-3
We move all terms to the left:
-(1/2)j-(-3)=0
Domain of the equation: 2)j!=0
j!=0/1
j!=0
j∈R
We add all the numbers together, and all the variables
-(+1/2)j-(-3)=0
We add all the numbers together, and all the variables
-(+1/2)j+3=0
We multiply parentheses
-j^2+3=0
We add all the numbers together, and all the variables
-1j^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:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{3}}{2*-1}=\frac{0-2\sqrt{3}}{-2} =-\frac{2\sqrt{3}}{-2} =-\frac{\sqrt{3}}{-1} $
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{3}}{2*-1}=\frac{0+2\sqrt{3}}{-2} =\frac{2\sqrt{3}}{-2} =\frac{\sqrt{3}}{-1} $

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