x+0.04+x+x^2=1

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

x+0.04+x+x^2=1

We move all terms to the left:

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

We add all the numbers together, and all the variables

x^2+2x-0.96=0

a = 1; b = 2; c = -0.96;
Δ = b2-4ac
Δ = 22-4·1·(-0.96)
Δ = 7.84
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{-(2)-\sqrt{7.84}}{2*1}=\frac{-2-\sqrt{7.84}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+\sqrt{7.84}}{2*1}=\frac{-2+\sqrt{7.84}}{2}
$