5t^2+100-100t=0

Solution for 5t^2+100-100t=0 equation:

5t^2+100-100t=0

a = 5; b = -100; c = +100;
Δ = b2-4ac
Δ = -1002-4·5·100
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{-(-100)-40\sqrt{5}}{2*5}=\frac{100-40\sqrt{5}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-100)+40\sqrt{5}}{2*5}=\frac{100+40\sqrt{5}}{10}
$