320=2x^2+4x

Solution for 320=2x^2+4x equation:

320=2x^2+4x

We move all terms to the left:

320-(2x^2+4x)=0

We get rid of parentheses

-2x^2-4x+320=0

a = -2; b = -4; c = +320;
Δ = b2-4ac
Δ = -42-4·(-2)·320
Δ = 2576
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:

$\sqrt{\Delta}=\sqrt{2576}=\sqrt{16*161}=\sqrt{16}*\sqrt{161}=4\sqrt{161}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{161}}{2*-2}=\frac{4-4\sqrt{161}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{161}}{2*-2}=\frac{4+4\sqrt{161}}{-4}
$