4(-4t^2+t+60)=0

Solution for 4(-4t^2+t+60)=0 equation:

4(-4t^2+t+60)=0

We multiply parentheses

-16t^2+4t+240=0

a = -16; b = 4; c = +240;
Δ = b2-4ac
Δ = 42-4·(-16)·240
Δ = 15376
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{15376}=124$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-124}{2*-16}=\frac{-128}{-32}
=+4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+124}{2*-16}=\frac{120}{-32}
=-3+3/4
$