0.25t^2+2t-20=0

Solution for 0.25t^2+2t-20=0 equation:

0.25t^2+2t-20=0

a = 0.25; b = 2; c = -20;
Δ = b2-4ac
Δ = 22-4·0.25·(-20)
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{6}}{2*0.25}=\frac{-2-2\sqrt{6}}{0.5}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{6}}{2*0.25}=\frac{-2+2\sqrt{6}}{0.5}
$