15=-16t2+20t+50

Solution for 15=-16t2+20t+50 equation:

15=-16t^2+20t+50

We move all terms to the left:

15-(-16t^2+20t+50)=0

We get rid of parentheses

16t^2-20t-50+15=0

We add all the numbers together, and all the variables

16t^2-20t-35=0

a = 16; b = -20; c = -35;
Δ = b2-4ac
Δ = -202-4·16·(-35)
Δ = 2640
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:

$\sqrt{\Delta}=\sqrt{2640}=\sqrt{16*165}=\sqrt{16}*\sqrt{165}=4\sqrt{165}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{165}}{2*16}=\frac{20-4\sqrt{165}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{165}}{2*16}=\frac{20+4\sqrt{165}}{32}
$