0=50t^2+100t-130

Solution for 0=50t^2+100t-130 equation:

0=50t^2+100t-130

We move all terms to the left:

0-(50t^2+100t-130)=0

We add all the numbers together, and all the variables

-(50t^2+100t-130)=0

We get rid of parentheses

-50t^2-100t+130=0

a = -50; b = -100; c = +130;
Δ = b2-4ac
Δ = -1002-4·(-50)·130
Δ = 36000
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:

$\sqrt{\Delta}=\sqrt{36000}=\sqrt{3600*10}=\sqrt{3600}*\sqrt{10}=60\sqrt{10}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-100)-60\sqrt{10}}{2*-50}=\frac{100-60\sqrt{10}}{-100}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-100)+60\sqrt{10}}{2*-50}=\frac{100+60\sqrt{10}}{-100}
$