1000=50t^2+100t+80

Solution for 1000=50t^2+100t+80 equation:

1000=50t^2+100t+80

We move all terms to the left:

1000-(50t^2+100t+80)=0

We get rid of parentheses

-50t^2-100t-80+1000=0

We add all the numbers together, and all the variables

-50t^2-100t+920=0

a = -50; b = -100; c = +920;
Δ = b2-4ac
Δ = -1002-4·(-50)·920
Δ = 194000
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:

$\sqrt{\Delta}=\sqrt{194000}=\sqrt{400*485}=\sqrt{400}*\sqrt{485}=20\sqrt{485}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-100)-20\sqrt{485}}{2*-50}=\frac{100-20\sqrt{485}}{-100}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-100)+20\sqrt{485}}{2*-50}=\frac{100+20\sqrt{485}}{-100}
$