-16t2+46t+6=0

Solution for -16t2+46t+6=0 equation:

-16t^2+46t+6=0

a = -16; b = 46; c = +6;
Δ = b2-4ac
Δ = 462-4·(-16)·6
Δ = 2500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2500}=50$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(46)-50}{2*-16}=\frac{-96}{-32}
=+3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(46)+50}{2*-16}=\frac{4}{-32}
=-1/8
$