j-6=1/2j+5

Solution for j-6=1/2j+5 equation:

j-6=1/2j+5

We move all terms to the left:

j-6-(1/2j+5)=0


Domain of the equation: 2j+5)!=0

j∈R


We get rid of parentheses

j-1/2j-5-6=0

We multiply all the terms by the denominator


j*2j-5*2j-6*2j-1=0

Wy multiply elements

2j^2-10j-12j-1=0

We add all the numbers together, and all the variables

2j^2-22j-1=0

a = 2; b = -22; c = -1;
Δ = b2-4ac
Δ = -222-4·2·(-1)
Δ = 492
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:

$\sqrt{\Delta}=\sqrt{492}=\sqrt{4*123}=\sqrt{4}*\sqrt{123}=2\sqrt{123}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{123}}{2*2}=\frac{22-2\sqrt{123}}{4}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{123}}{2*2}=\frac{22+2\sqrt{123}}{4}
$