-16t^2+20t+6=10

Solution for -16t^2+20t+6=10 equation:

-16t^2+20t+6=10

We move all terms to the left:

-16t^2+20t+6-(10)=0

We add all the numbers together, and all the variables

-16t^2+20t-4=0

a = -16; b = 20; c = -4;
Δ = b2-4ac
Δ = 202-4·(-16)·(-4)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-12}{2*-16}=\frac{-32}{-32}
=1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+12}{2*-16}=\frac{-8}{-32}
=1/4
$