Y=5t-10t^2+10

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

=5Y-10Y^2+10

We move all terms to the left:

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

We get rid of parentheses

10Y^2-5Y-10=0

a = 10; b = -5; c = -10;
Δ = b2-4ac
Δ = -52-4·10·(-10)
Δ = 425
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:

$\sqrt{\Delta}=\sqrt{425}=\sqrt{25*17}=\sqrt{25}*\sqrt{17}=5\sqrt{17}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5\sqrt{17}}{2*10}=\frac{5-5\sqrt{17}}{20}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5\sqrt{17}}{2*10}=\frac{5+5\sqrt{17}}{20}
$