0=-16x^2+160x+1200

Solution for 0=-16x^2+160x+1200 equation:

0=-16x^2+160x+1200

We move all terms to the left:

0-(-16x^2+160x+1200)=0

We add all the numbers together, and all the variables

-(-16x^2+160x+1200)=0

We get rid of parentheses

16x^2-160x-1200=0

a = 16; b = -160; c = -1200;
Δ = b2-4ac
Δ = -1602-4·16·(-1200)
Δ = 102400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{102400}=320$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-320}{2*16}=\frac{-160}{32}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+320}{2*16}=\frac{480}{32}
=15
$