338=3x^2+17x-160

Solution for 338=3x^2+17x-160 equation:

338=3x^2+17x-160

We move all terms to the left:

338-(3x^2+17x-160)=0

We get rid of parentheses

-3x^2-17x+160+338=0

We add all the numbers together, and all the variables

-3x^2-17x+498=0

a = -3; b = -17; c = +498;
Δ = b2-4ac
Δ = -172-4·(-3)·498
Δ = 6265
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{-(-17)-\sqrt{6265}}{2*-3}=\frac{17-\sqrt{6265}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+\sqrt{6265}}{2*-3}=\frac{17+\sqrt{6265}}{-6}
$