2x(3x+17)=320

Solution for 2x(3x+17)=320 equation:

2x(3x+17)=320

We move all terms to the left:

2x(3x+17)-(320)=0

We multiply parentheses

6x^2+34x-320=0

a = 6; b = 34; c = -320;
Δ = b2-4ac
Δ = 342-4·6·(-320)
Δ = 8836
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{8836}=94$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-94}{2*6}=\frac{-128}{12}
=-10+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+94}{2*6}=\frac{60}{12}
=5
$