17=2(3x+1)x

Solution for 17=2(3x+1)x equation:

17=2(3x+1)x

We move all terms to the left:

17-(2(3x+1)x)=0


We calculate terms in parentheses: -(2(3x+1)x), so:

2(3x+1)x

We multiply parentheses

6x^2+2x

Back to the equation:

-(6x^2+2x)


We get rid of parentheses

-6x^2-2x+17=0

a = -6; b = -2; c = +17;
Δ = b2-4ac
Δ = -22-4·(-6)·17
Δ = 412
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{412}=\sqrt{4*103}=\sqrt{4}*\sqrt{103}=2\sqrt{103}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{103}}{2*-6}=\frac{2-2\sqrt{103}}{-12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{103}}{2*-6}=\frac{2+2\sqrt{103}}{-12}
$