Y=-5t^2-4t+120

Solution for Y=-5t^2-4t+120 equation:

=-5Y^2-4Y+120

We move all terms to the left:

-(-5Y^2-4Y+120)=0

We get rid of parentheses

5Y^2+4Y-120=0

a = 5; b = 4; c = -120;
Δ = b2-4ac
Δ = 42-4·5·(-120)
Δ = 2416
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:

$\sqrt{\Delta}=\sqrt{2416}=\sqrt{16*151}=\sqrt{16}*\sqrt{151}=4\sqrt{151}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{151}}{2*5}=\frac{-4-4\sqrt{151}}{10}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{151}}{2*5}=\frac{-4+4\sqrt{151}}{10}
$