.228075=4((10)^-05)x+0.013

Solution for .228075=4((10)^-05)x+0.013 equation:

.228075=4((10)^-05)x+0.013

We move all terms to the left:

.228075-(4((10)^-05)x+0.013)=0

We add all the numbers together, and all the variables

-(4(10^-05)x+0.013)+0.228075=0


We calculate terms in parentheses: -(4(10^-05)x+0.013), so:

4(10^-05)x+0.013

We multiply parentheses

40x^2-20x+0.013

Back to the equation:

-(40x^2-20x+0.013)


We get rid of parentheses

-40x^2+20x-0.013+0.228075=0

We add all the numbers together, and all the variables

-40x^2+20x+0.215075=0

a = -40; b = 20; c = +0.215075;
Δ = b2-4ac
Δ = 202-4·(-40)·0.215075
Δ = 434.412
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-\sqrt{434.412}}{2*-40}=\frac{-20-\sqrt{434.412}}{-80}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+\sqrt{434.412}}{2*-40}=\frac{-20+\sqrt{434.412}}{-80}
$