t=-16t^2+300

Solution for t=-16t^2+300 equation:

t=-16t^2+300

We move all terms to the left:

t-(-16t^2+300)=0

We get rid of parentheses

16t^2+t-300=0

a = 16; b = 1; c = -300;
Δ = b2-4ac
Δ = 12-4·16·(-300)
Δ = 19201
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{19201}}{2*16}=\frac{-1-\sqrt{19201}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{19201}}{2*16}=\frac{-1+\sqrt{19201}}{32}
$