16t^2-188t+92=0

Solution for 16t^2-188t+92=0 equation:

16t^2-188t+92=0

a = 16; b = -188; c = +92;
Δ = b2-4ac
Δ = -1882-4·16·92
Δ = 29456
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:

$\sqrt{\Delta}=\sqrt{29456}=\sqrt{16*1841}=\sqrt{16}*\sqrt{1841}=4\sqrt{1841}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-188)-4\sqrt{1841}}{2*16}=\frac{188-4\sqrt{1841}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-188)+4\sqrt{1841}}{2*16}=\frac{188+4\sqrt{1841}}{32}
$