289=j^2

Solution for 289=j^2 equation:

289=j^2

We move all terms to the left:

289-(j^2)=0

We add all the numbers together, and all the variables

-1j^2+289=0

a = -1; b = 0; c = +289;
Δ = b2-4ac
Δ = 02-4·(-1)·289
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1156}=34$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-34}{2*-1}=\frac{-34}{-2}
=+17
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+34}{2*-1}=\frac{34}{-2}
=-17
$