0.12+0.18+(0.6*x^2)+(1-x)+0.1=1

Solution for 0.12+0.18+(0.6*x^2)+(1-x)+0.1=1 equation:

0.12+0.18+(0.6*x^2)+(1-x)+0.1=1

We move all terms to the left:

0.12+0.18+(0.6*x^2)+(1-x)+0.1-(1)=0

We add all the numbers together, and all the variables

(0.6*x^2)+(-1x+1)+0.12+0.18+0.1-1=0

We add all the numbers together, and all the variables

(0.6*x^2)+(-1x+1)-0.6=0

We get rid of parentheses

0.6*x^2-1x+1-0.6=0

We add all the numbers together, and all the variables

0.6x^2-1x+0.4=0

a = 0.6; b = -1; c = +0.4;
Δ = b2-4ac
Δ = -12-4·0.6·0.4
Δ = 0.04
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{-(-1)-\sqrt{0.04}}{2*0.6}=\frac{1-\sqrt{0.04}}{1.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{0.04}}{2*0.6}=\frac{1+\sqrt{0.04}}{1.2}
$