0=16t^2+60t+40

Solution for 0=16t^2+60t+40 equation:

0=16t^2+60t+40

We move all terms to the left:

0-(16t^2+60t+40)=0

We add all the numbers together, and all the variables

-(16t^2+60t+40)=0

We get rid of parentheses

-16t^2-60t-40=0

a = -16; b = -60; c = -40;
Δ = b2-4ac
Δ = -602-4·(-16)·(-40)
Δ = 1040
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:

$\sqrt{\Delta}=\sqrt{1040}=\sqrt{16*65}=\sqrt{16}*\sqrt{65}=4\sqrt{65}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-60)-4\sqrt{65}}{2*-16}=\frac{60-4\sqrt{65}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-60)+4\sqrt{65}}{2*-16}=\frac{60+4\sqrt{65}}{-32}
$