2/36j+9=3j+7

Solution for 2/36j+9=3j+7 equation:

2/36j+9=3j+7

We move all terms to the left:

2/36j+9-(3j+7)=0


Domain of the equation: 36j!=0

j!=0/36

j!=0

j∈R


We get rid of parentheses

2/36j-3j-7+9=0

We multiply all the terms by the denominator


-3j*36j-7*36j+9*36j+2=0

Wy multiply elements

-108j^2-252j+324j+2=0

We add all the numbers together, and all the variables

-108j^2+72j+2=0

a = -108; b = 72; c = +2;
Δ = b2-4ac
Δ = 722-4·(-108)·2
Δ = 6048
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:

$\sqrt{\Delta}=\sqrt{6048}=\sqrt{144*42}=\sqrt{144}*\sqrt{42}=12\sqrt{42}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(72)-12\sqrt{42}}{2*-108}=\frac{-72-12\sqrt{42}}{-216}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(72)+12\sqrt{42}}{2*-108}=\frac{-72+12\sqrt{42}}{-216}
$