.55^2+2(.55)(b)+b^2=0

Solution for .55^2+2(.55)(b)+b^2=0 equation:

.55^2+2(.55)(b)+b^2=0

We add all the numbers together, and all the variables

b^2+2(0.55)b+.55^2=0

We add all the numbers together, and all the variables

b^2+2(0.55)b+0.3025=0

We multiply parentheses

b^2+1.1b+0.3025=0

a = 1; b = 1.1; c = +0.3025;
Δ = b2-4ac
Δ = 1.12-4·1·0.3025
Δ = 2.2204460492503E-16
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{-(1.1)-\sqrt{2.2204460492503E-16}}{2*1}=\frac{-1.1-\sqrt{2.2204460492503E-16}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.1)+\sqrt{2.2204460492503E-16}}{2*1}=\frac{-1.1+\sqrt{2.2204460492503E-16}}{2}
$