j-16=1/2j+3

Solution for j-16=1/2j+3 equation:

j-16=1/2j+3

We move all terms to the left:

j-16-(1/2j+3)=0


Domain of the equation: 2j+3)!=0

j∈R


We get rid of parentheses

j-1/2j-3-16=0

We multiply all the terms by the denominator


j*2j-3*2j-16*2j-1=0

Wy multiply elements

2j^2-6j-32j-1=0

We add all the numbers together, and all the variables

2j^2-38j-1=0

a = 2; b = -38; c = -1;
Δ = b2-4ac
Δ = -382-4·2·(-1)
Δ = 1452
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:

$\sqrt{\Delta}=\sqrt{1452}=\sqrt{484*3}=\sqrt{484}*\sqrt{3}=22\sqrt{3}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-22\sqrt{3}}{2*2}=\frac{38-22\sqrt{3}}{4}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+22\sqrt{3}}{2*2}=\frac{38+22\sqrt{3}}{4}
$