2j^2+2j=J^2-156j

Solution for 2j^2+2j=J^2-156j equation:

2j^2+2j=^2-156j

We move all terms to the left:

2j^2+2j-(^2-156j)=0

We get rid of parentheses

2j^2+2j+156j-^2=0

We add all the numbers together, and all the variables

2j^2+158j=0

a = 2; b = 158; c = 0;
Δ = b2-4ac
Δ = 1582-4·2·0
Δ = 24964
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{24964}=158$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(158)-158}{2*2}=\frac{-316}{4}
=-79
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(158)+158}{2*2}=\frac{0}{4}
=0
$