0=16t^2+50t+10

Solution for 0=16t^2+50t+10 equation:

0=16t^2+50t+10

We move all terms to the left:

0-(16t^2+50t+10)=0

We add all the numbers together, and all the variables

-(16t^2+50t+10)=0

We get rid of parentheses

-16t^2-50t-10=0

a = -16; b = -50; c = -10;
Δ = b2-4ac
Δ = -502-4·(-16)·(-10)
Δ = 1860
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:

$\sqrt{\Delta}=\sqrt{1860}=\sqrt{4*465}=\sqrt{4}*\sqrt{465}=2\sqrt{465}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-2\sqrt{465}}{2*-16}=\frac{50-2\sqrt{465}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+2\sqrt{465}}{2*-16}=\frac{50+2\sqrt{465}}{-32}
$