b^2=0.25

Solution for b^2=0.25 equation:

b^2=0.25

We move all terms to the left:

b^2-(0.25)=0

We add all the numbers together, and all the variables

b^2-0.25=0

a = 1; b = 0; c = -0.25;
Δ = b2-4ac
Δ = 02-4·1·(-0.25)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-1}{2*1}=\frac{-1}{2}
=-1/2
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+1}{2*1}=\frac{1}{2}
=1/2
$