0=-16x^2-48x+160

Solution for 0=-16x^2-48x+160 equation:

0=-16x^2-48x+160

We move all terms to the left:

0-(-16x^2-48x+160)=0

We add all the numbers together, and all the variables

-(-16x^2-48x+160)=0

We get rid of parentheses

16x^2+48x-160=0

a = 16; b = 48; c = -160;
Δ = b2-4ac
Δ = 482-4·16·(-160)
Δ = 12544
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12544}=112$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-112}{2*16}=\frac{-160}{32}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+112}{2*16}=\frac{64}{32}
=2
$