0.25t^2+20t=10000

Solution for 0.25t^2+20t=10000 equation:

0.25t^2+20t=10000

We move all terms to the left:

0.25t^2+20t-(10000)=0

a = 0.25; b = 20; c = -10000;
Δ = b2-4ac
Δ = 202-4·0.25·(-10000)
Δ = 10400
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:

$\sqrt{\Delta}=\sqrt{10400}=\sqrt{400*26}=\sqrt{400}*\sqrt{26}=20\sqrt{26}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20\sqrt{26}}{2*0.25}=\frac{-20-20\sqrt{26}}{0.5}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20\sqrt{26}}{2*0.25}=\frac{-20+20\sqrt{26}}{0.5}
$