5j^2=20

Solution for 5j^2=20 equation:

5j^2=20

We move all terms to the left:

5j^2-(20)=0

a = 5; b = 0; c = -20;
Δ = b2-4ac
Δ = 02-4·5·(-20)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20}{2*5}=\frac{-20}{10}
=-2
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20}{2*5}=\frac{20}{10}
=2
$