x^2+0.4x+0.04=1

Solution for x^2+0.4x+0.04=1 equation:

x^2+0.4x+0.04=1

We move all terms to the left:

x^2+0.4x+0.04-(1)=0

We add all the numbers together, and all the variables

x^2+0.4x-0.96=0

a = 1; b = 0.4; c = -0.96;
Δ = b2-4ac
Δ = 0.42-4·1·(-0.96)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0.4)-2}{2*1}=\frac{-2.4}{2}
=-1+0.4/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.4)+2}{2*1}=\frac{1.6}{2}
=1/1.25
$