-16t^2+20t+5=10

Solution for -16t^2+20t+5=10 equation:

-16t^2+20t+5=10

We move all terms to the left:

-16t^2+20t+5-(10)=0

We add all the numbers together, and all the variables

-16t^2+20t-5=0

a = -16; b = 20; c = -5;
Δ = b2-4ac
Δ = 202-4·(-16)·(-5)
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{5}}{2*-16}=\frac{-20-4\sqrt{5}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{5}}{2*-16}=\frac{-20+4\sqrt{5}}{-32}
$