40x^2+4x+0.2=1

Solution for 40x^2+4x+0.2=1 equation:

40x^2+4x+0.2=1

We move all terms to the left:

40x^2+4x+0.2-(1)=0

We add all the numbers together, and all the variables

40x^2+4x-0.8=0

a = 40; b = 4; c = -0.8;
Δ = b2-4ac
Δ = 42-4·40·(-0.8)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-12}{2*40}=\frac{-16}{80}
=-1/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+12}{2*40}=\frac{8}{80}
=1/10
$