16t^2+60t+4=0

Solution for 16t^2+60t+4=0 equation:

16t^2+60t+4=0

a = 16; b = 60; c = +4;
Δ = b2-4ac
Δ = 602-4·16·4
Δ = 3344
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:

$\sqrt{\Delta}=\sqrt{3344}=\sqrt{16*209}=\sqrt{16}*\sqrt{209}=4\sqrt{209}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-4\sqrt{209}}{2*16}=\frac{-60-4\sqrt{209}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+4\sqrt{209}}{2*16}=\frac{-60+4\sqrt{209}}{32}
$