-t^2+120t=1600

Solution for -t^2+120t=1600 equation:

-t^2+120t=1600

We move all terms to the left:

-t^2+120t-(1600)=0

We add all the numbers together, and all the variables

-1t^2+120t-1600=0

a = -1; b = 120; c = -1600;
Δ = b2-4ac
Δ = 1202-4·(-1)·(-1600)
Δ = 8000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{8000}=\sqrt{1600*5}=\sqrt{1600}*\sqrt{5}=40\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(120)-40\sqrt{5}}{2*-1}=\frac{-120-40\sqrt{5}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(120)+40\sqrt{5}}{2*-1}=\frac{-120+40\sqrt{5}}{-2}
$