Y=2t^2-16t+20

Solution for Y=2t^2-16t+20 equation:

=2Y^2-16Y+20

We move all terms to the left:

-(2Y^2-16Y+20)=0

We get rid of parentheses

-2Y^2+16Y-20=0

a = -2; b = 16; c = -20;
Δ = b2-4ac
Δ = 162-4·(-2)·(-20)
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{6}}{2*-2}=\frac{-16-4\sqrt{6}}{-4}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{6}}{2*-2}=\frac{-16+4\sqrt{6}}{-4}
$