4t^2+20t-100=0

Solution for 4t^2+20t-100=0 equation:

4t^2+20t-100=0

a = 4; b = 20; c = -100;
Δ = b2-4ac
Δ = 202-4·4·(-100)
Δ = 2000
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:

$\sqrt{\Delta}=\sqrt{2000}=\sqrt{400*5}=\sqrt{400}*\sqrt{5}=20\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20\sqrt{5}}{2*4}=\frac{-20-20\sqrt{5}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20\sqrt{5}}{2*4}=\frac{-20+20\sqrt{5}}{8}
$