7-3j+10=-7/9j

Solution for 7-3j+10=-7/9j equation:

7-3j+10=-7/9j

We move all terms to the left:

7-3j+10-(-7/9j)=0


Domain of the equation: 9j)!=0

j!=0/1

j!=0

j∈R


We add all the numbers together, and all the variables

-3j-(-7/9j)+17=0

We get rid of parentheses

-3j+7/9j+17=0

We multiply all the terms by the denominator


-3j*9j+17*9j+7=0

Wy multiply elements

-27j^2+153j+7=0

a = -27; b = 153; c = +7;
Δ = b2-4ac
Δ = 1532-4·(-27)·7
Δ = 24165
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:

$\sqrt{\Delta}=\sqrt{24165}=\sqrt{9*2685}=\sqrt{9}*\sqrt{2685}=3\sqrt{2685}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(153)-3\sqrt{2685}}{2*-27}=\frac{-153-3\sqrt{2685}}{-54}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(153)+3\sqrt{2685}}{2*-27}=\frac{-153+3\sqrt{2685}}{-54}
$