20=-0.6t^2+24t

Solution for 20=-0.6t^2+24t equation:

20=-0.6t^2+24t

We move all terms to the left:

20-(-0.6t^2+24t)=0

We get rid of parentheses

0.6t^2-24t+20=0

a = 0.6; b = -24; c = +20;
Δ = b2-4ac
Δ = -242-4·0.6·20
Δ = 528
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:

$\sqrt{\Delta}=\sqrt{528}=\sqrt{16*33}=\sqrt{16}*\sqrt{33}=4\sqrt{33}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-4\sqrt{33}}{2*0.6}=\frac{24-4\sqrt{33}}{1.2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+4\sqrt{33}}{2*0.6}=\frac{24+4\sqrt{33}}{1.2}
$