-16t^2+226t+105=0

Solution for -16t^2+226t+105=0 equation:

-16t^2+226t+105=0

a = -16; b = 226; c = +105;
Δ = b2-4ac
Δ = 2262-4·(-16)·105
Δ = 57796
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:

$\sqrt{\Delta}=\sqrt{57796}=\sqrt{4*14449}=\sqrt{4}*\sqrt{14449}=2\sqrt{14449}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(226)-2\sqrt{14449}}{2*-16}=\frac{-226-2\sqrt{14449}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(226)+2\sqrt{14449}}{2*-16}=\frac{-226+2\sqrt{14449}}{-32}
$