Y=-5t2+30t+35

Solution for Y=-5t2+30t+35 equation:

=-5Y^2+30Y+35

We move all terms to the left:

-(-5Y^2+30Y+35)=0

We get rid of parentheses

5Y^2-30Y-35=0

a = 5; b = -30; c = -35;
Δ = b2-4ac
Δ = -302-4·5·(-35)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-40}{2*5}=\frac{-10}{10}
=-1
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+40}{2*5}=\frac{70}{10}
=7
$