Y=20+10t-5t^2

Solution for Y=20+10t-5t^2 equation:

=20+10Y-5Y^2

We move all terms to the left:

-(20+10Y-5Y^2)=0

We get rid of parentheses

5Y^2-10Y-20=0

a = 5; b = -10; c = -20;
Δ = b2-4ac
Δ = -102-4·5·(-20)
Δ = 500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{500}=\sqrt{100*5}=\sqrt{100}*\sqrt{5}=10\sqrt{5}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10\sqrt{5}}{2*5}=\frac{10-10\sqrt{5}}{10}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10\sqrt{5}}{2*5}=\frac{10+10\sqrt{5}}{10}
$