0=(-16t^2)+300

Solution for 0=(-16t^2)+300 equation:

0=(-16t^2)+300

We move all terms to the left:

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

We add all the numbers together, and all the variables

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


We calculate terms in parentheses: -((-16t^2)+300), so:

(-16t^2)+300

We get rid of parentheses

-16t^2+300

Back to the equation:

-(-16t^2+300)


We get rid of parentheses

16t^2-300=0

a = 16; b = 0; c = -300;
Δ = b2-4ac
Δ = 02-4·16·(-300)
Δ = 19200
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:

$\sqrt{\Delta}=\sqrt{19200}=\sqrt{6400*3}=\sqrt{6400}*\sqrt{3}=80\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-80\sqrt{3}}{2*16}=\frac{0-80\sqrt{3}}{32}
=-\frac{80\sqrt{3}}{32}
=-\frac{5\sqrt{3}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+80\sqrt{3}}{2*16}=\frac{0+80\sqrt{3}}{32}
=\frac{80\sqrt{3}}{32}
=\frac{5\sqrt{3}}{2}
$