1+x=9x-6x(x+3.5)+0.2

Solution for 1+x=9x-6x(x+3.5)+0.2 equation:

1+x=9x-6x(x+3.5)+0.2

We move all terms to the left:

1+x-(9x-6x(x+3.5)+0.2)=0


We calculate terms in parentheses: -(9x-6x(x+3.5)+0.2), so:

9x-6x(x+3.5)+0.2

We multiply parentheses

-6x^2+9x-21x+0.2

We add all the numbers together, and all the variables

-6x^2-12x+0.2

Back to the equation:

-(-6x^2-12x+0.2)


We get rid of parentheses

6x^2+12x+x-0.2+1=0

We add all the numbers together, and all the variables

6x^2+13x+0.8=0

a = 6; b = 13; c = +0.8;
Δ = b2-4ac
Δ = 132-4·6·0.8
Δ = 149.8
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{-(13)-\sqrt{149.8}}{2*6}=\frac{-13-\sqrt{149.8}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{149.8}}{2*6}=\frac{-13+\sqrt{149.8}}{12}
$