16t^2+72t+5=0

Solution for 16t^2+72t+5=0 equation:

16t^2+72t+5=0

a = 16; b = 72; c = +5;
Δ = b2-4ac
Δ = 722-4·16·5
Δ = 4864
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:

$\sqrt{\Delta}=\sqrt{4864}=\sqrt{256*19}=\sqrt{256}*\sqrt{19}=16\sqrt{19}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(72)-16\sqrt{19}}{2*16}=\frac{-72-16\sqrt{19}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(72)+16\sqrt{19}}{2*16}=\frac{-72+16\sqrt{19}}{32}
$