35=2(x+3)(3x+1)+2x

Solution for 35=2(x+3)(3x+1)+2x equation:

35=2(x+3)(3x+1)+2x

We move all terms to the left:

35-(2(x+3)(3x+1)+2x)=0

We multiply parentheses ..

-(2(+3x^2+x+9x+3)+2x)+35=0


We calculate terms in parentheses: -(2(+3x^2+x+9x+3)+2x), so:

2(+3x^2+x+9x+3)+2x

We multiply parentheses

6x^2+2x+18x+2x+6

We add all the numbers together, and all the variables

6x^2+22x+6

Back to the equation:

-(6x^2+22x+6)


We get rid of parentheses

-6x^2-22x-6+35=0

We add all the numbers together, and all the variables

-6x^2-22x+29=0

a = -6; b = -22; c = +29;
Δ = b2-4ac
Δ = -222-4·(-6)·29
Δ = 1180
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1180}=\sqrt{4*295}=\sqrt{4}*\sqrt{295}=2\sqrt{295}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{295}}{2*-6}=\frac{22-2\sqrt{295}}{-12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{295}}{2*-6}=\frac{22+2\sqrt{295}}{-12}
$