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

Solution for 0=-16x^2+128x+-240 equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We use the square of the difference formula

-(-16x^2+128x-240)=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 = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-128)-32}{2*16}=\frac{96}{32}
=3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-128)+32}{2*16}=\frac{160}{32}
=5
$