0=-16t^2+150+50

Solution for 0=-16t^2+150+50 equation:

0=-16t^2+150+50

We move all terms to the left:

0-(-16t^2+150+50)=0

We add all the numbers together, and all the variables

-(-16t^2+150+50)=0

We get rid of parentheses

16t^2-150-50=0

We add all the numbers together, and all the variables

16t^2-200=0

a = 16; b = 0; c = -200;
Δ = b2-4ac
Δ = 02-4·16·(-200)
Δ = 12800
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:

$\sqrt{\Delta}=\sqrt{12800}=\sqrt{6400*2}=\sqrt{6400}*\sqrt{2}=80\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-80\sqrt{2}}{2*16}=\frac{0-80\sqrt{2}}{32}
=-\frac{80\sqrt{2}}{32}
=-\frac{5\sqrt{2}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+80\sqrt{2}}{2*16}=\frac{0+80\sqrt{2}}{32}
=\frac{80\sqrt{2}}{32}
=\frac{5\sqrt{2}}{2}
$