4j^2=25

Solution for 4j^2=25 equation:

4j^2=25

We move all terms to the left:

4j^2-(25)=0

a = 4; b = 0; c = -25;
Δ = b2-4ac
Δ = 02-4·4·(-25)
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*4}=\frac{-20}{8}
=-2+1/2
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20}{2*4}=\frac{20}{8}
=2+1/2
$