.25=b^2-4

Solution for .25=b^2-4 equation:

.25=b^2-4

We move all terms to the left:

.25-(b^2-4)=0

We add all the numbers together, and all the variables

-(b^2-4)+0.25=0

We get rid of parentheses

-b^2+4+0.25=0

We add all the numbers together, and all the variables

-1b^2+4.25=0

a = -1; b = 0; c = +4.25;
Δ = b2-4ac
Δ = 02-4·(-1)·4.25
Δ = 17
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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