16t^2+80t+3*5=0

Solution for 16t^2+80t+3*5=0 equation:

16t^2+80t+3*5=0

We add all the numbers together, and all the variables

16t^2+80t+15=0

a = 16; b = 80; c = +15;
Δ = b2-4ac
Δ = 802-4·16·15
Δ = 5440
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:

$\sqrt{\Delta}=\sqrt{5440}=\sqrt{64*85}=\sqrt{64}*\sqrt{85}=8\sqrt{85}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-8\sqrt{85}}{2*16}=\frac{-80-8\sqrt{85}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+8\sqrt{85}}{2*16}=\frac{-80+8\sqrt{85}}{32}
$