(1.1)^2+(b)^2=(1.6)^2

Solution for (1.1)^2+(b)^2=(1.6)^2 equation:

(1.1)^2+(b)^2=(1.6)^2

We move all terms to the left:

(1.1)^2+(b)^2-((1.6)^2)=0

We add all the numbers together, and all the variables

b^2-1.35=0

a = 1; b = 0; c = -1.35;
Δ = b2-4ac
Δ = 02-4·1·(-1.35)
Δ = 5.4
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{5.4}}{2*1}=\frac{0-\sqrt{5.4}}{2}
=-\frac{\sqrt{}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{5.4}}{2*1}=\frac{0+\sqrt{5.4}}{2}
=\frac{\sqrt{}}{2}
$