-t^2+100t-50=0

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

-t^2+100t-50=0

We add all the numbers together, and all the variables

-1t^2+100t-50=0

a = -1; b = 100; c = -50;
Δ = b2-4ac
Δ = 1002-4·(-1)·(-50)
Δ = 9800
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:

$\sqrt{\Delta}=\sqrt{9800}=\sqrt{4900*2}=\sqrt{4900}*\sqrt{2}=70\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-70\sqrt{2}}{2*-1}=\frac{-100-70\sqrt{2}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+70\sqrt{2}}{2*-1}=\frac{-100+70\sqrt{2}}{-2}
$