240=16x^2+128x

Solution for 240=16x^2+128x equation:

240=16x^2+128x

We move all terms to the left:

240-(16x^2+128x)=0

We get rid of parentheses

-16x^2-128x+240=0

a = -16; b = -128; c = +240;
Δ = b2-4ac
Δ = -1282-4·(-16)·240
Δ = 31744
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:

$\sqrt{\Delta}=\sqrt{31744}=\sqrt{1024*31}=\sqrt{1024}*\sqrt{31}=32\sqrt{31}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-128)-32\sqrt{31}}{2*-16}=\frac{128-32\sqrt{31}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-128)+32\sqrt{31}}{2*-16}=\frac{128+32\sqrt{31}}{-32}
$