-16t^2+20+300=5

Solution for -16t^2+20+300=5 equation:

-16t^2+20+300=5

We move all terms to the left:

-16t^2+20+300-(5)=0

We add all the numbers together, and all the variables

-16t^2+315=0

a = -16; b = 0; c = +315;
Δ = b2-4ac
Δ = 02-4·(-16)·315
Δ = 20160
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:

$\sqrt{\Delta}=\sqrt{20160}=\sqrt{576*35}=\sqrt{576}*\sqrt{35}=24\sqrt{35}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24\sqrt{35}}{2*-16}=\frac{0-24\sqrt{35}}{-32}
=-\frac{24\sqrt{35}}{-32}
=-\frac{3\sqrt{35}}{-4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24\sqrt{35}}{2*-16}=\frac{0+24\sqrt{35}}{-32}
=\frac{24\sqrt{35}}{-32}
=\frac{3\sqrt{35}}{-4}
$