25q^2=4

Solution for 25q^2=4 equation:

25q^2=4

We move all terms to the left:

25q^2-(4)=0

a = 25; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·25·(-4)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20}{2*25}=\frac{-20}{50}
=-2/5
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20}{2*25}=\frac{20}{50}
=2/5
$