-16t^2+20t+6=3

Solution for -16t^2+20t+6=3 equation:

-16t^2+20t+6=3

We move all terms to the left:

-16t^2+20t+6-(3)=0

We add all the numbers together, and all the variables

-16t^2+20t+3=0

a = -16; b = 20; c = +3;
Δ = b2-4ac
Δ = 202-4·(-16)·3
Δ = 592
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:

$\sqrt{\Delta}=\sqrt{592}=\sqrt{16*37}=\sqrt{16}*\sqrt{37}=4\sqrt{37}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{37}}{2*-16}=\frac{-20-4\sqrt{37}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{37}}{2*-16}=\frac{-20+4\sqrt{37}}{-32}
$