-14t^2+56t+60=0

Solution for -14t^2+56t+60=0 equation:

-14t^2+56t+60=0

a = -14; b = 56; c = +60;
Δ = b2-4ac
Δ = 562-4·(-14)·60
Δ = 6496
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:

$\sqrt{\Delta}=\sqrt{6496}=\sqrt{16*406}=\sqrt{16}*\sqrt{406}=4\sqrt{406}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(56)-4\sqrt{406}}{2*-14}=\frac{-56-4\sqrt{406}}{-28}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(56)+4\sqrt{406}}{2*-14}=\frac{-56+4\sqrt{406}}{-28}
$