5t^2-20t-100=0

Solution for 5t^2-20t-100=0 equation:

5t^2-20t-100=0

a = 5; b = -20; c = -100;
Δ = b2-4ac
Δ = -202-4·5·(-100)
Δ = 2400
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:

$\sqrt{\Delta}=\sqrt{2400}=\sqrt{400*6}=\sqrt{400}*\sqrt{6}=20\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20\sqrt{6}}{2*5}=\frac{20-20\sqrt{6}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20\sqrt{6}}{2*5}=\frac{20+20\sqrt{6}}{10}
$