35(t)=-16t^2+35

Solution for 35(t)=-16t^2+35 equation:

35(t)=-16t^2+35

We move all terms to the left:

35(t)-(-16t^2+35)=0

We get rid of parentheses

16t^2+35t-35=0

a = 16; b = 35; c = -35;
Δ = b2-4ac
Δ = 352-4·16·(-35)
Δ = 3465
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{3465}=\sqrt{9*385}=\sqrt{9}*\sqrt{385}=3\sqrt{385}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-3\sqrt{385}}{2*16}=\frac{-35-3\sqrt{385}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+3\sqrt{385}}{2*16}=\frac{-35+3\sqrt{385}}{32}
$