Y=-5t2+50t-4

Solution for Y=-5t2+50t-4 equation:

=-5Y^2+50Y-4

We move all terms to the left:

-(-5Y^2+50Y-4)=0

We get rid of parentheses

5Y^2-50Y+4=0

a = 5; b = -50; c = +4;
Δ = b2-4ac
Δ = -502-4·5·4
Δ = 2420
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:

$\sqrt{\Delta}=\sqrt{2420}=\sqrt{484*5}=\sqrt{484}*\sqrt{5}=22\sqrt{5}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-22\sqrt{5}}{2*5}=\frac{50-22\sqrt{5}}{10}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+22\sqrt{5}}{2*5}=\frac{50+22\sqrt{5}}{10}
$