330=3x^2+17x

Solution for 330=3x^2+17x equation:

330=3x^2+17x

We move all terms to the left:

330-(3x^2+17x)=0

We get rid of parentheses

-3x^2-17x+330=0

a = -3; b = -17; c = +330;
Δ = b2-4ac
Δ = -172-4·(-3)·330
Δ = 4249
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{4249}}{2*-3}=\frac{17-\sqrt{4249}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+\sqrt{4249}}{2*-3}=\frac{17+\sqrt{4249}}{-6}
$