30(t)=-16t^2+30

Solution for 30(t)=-16t^2+30 equation:

30(t)=-16t^2+30

We move all terms to the left:

30(t)-(-16t^2+30)=0

We get rid of parentheses

16t^2+30t-30=0

a = 16; b = 30; c = -30;
Δ = b2-4ac
Δ = 302-4·16·(-30)
Δ = 2820
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:

$\sqrt{\Delta}=\sqrt{2820}=\sqrt{4*705}=\sqrt{4}*\sqrt{705}=2\sqrt{705}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-2\sqrt{705}}{2*16}=\frac{-30-2\sqrt{705}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+2\sqrt{705}}{2*16}=\frac{-30+2\sqrt{705}}{32}
$