50=-16t^2-225t+25

Solution for 50=-16t^2-225t+25 equation:

50=-16t^2-225t+25

We move all terms to the left:

50-(-16t^2-225t+25)=0

We get rid of parentheses

16t^2+225t-25+50=0

We add all the numbers together, and all the variables

16t^2+225t+25=0

a = 16; b = 225; c = +25;
Δ = b2-4ac
Δ = 2252-4·16·25
Δ = 49025
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:

$\sqrt{\Delta}=\sqrt{49025}=\sqrt{25*1961}=\sqrt{25}*\sqrt{1961}=5\sqrt{1961}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(225)-5\sqrt{1961}}{2*16}=\frac{-225-5\sqrt{1961}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(225)+5\sqrt{1961}}{2*16}=\frac{-225+5\sqrt{1961}}{32}
$