320=2x^2+3x

Solution for 320=2x^2+3x equation:

320=2x^2+3x

We move all terms to the left:

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

We get rid of parentheses

-2x^2-3x+320=0

a = -2; b = -3; c = +320;
Δ = b2-4ac
Δ = -32-4·(-2)·320
Δ = 2569
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{-(-3)-\sqrt{2569}}{2*-2}=\frac{3-\sqrt{2569}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{2569}}{2*-2}=\frac{3+\sqrt{2569}}{-4}
$