.5t^2-20t+2=0

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

.5t^2-20t+2=0

a = .5; b = -20; c = +2;
Δ = b2-4ac
Δ = -202-4·.5·2
Δ = 396
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:

$\sqrt{\Delta}=\sqrt{396}=\sqrt{36*11}=\sqrt{36}*\sqrt{11}=6\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-6\sqrt{11}}{2*.5}=\frac{20-6\sqrt{11}}{1}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+6\sqrt{11}}{2*.5}=\frac{20+6\sqrt{11}}{1}
$