16t^2+16t-60=0

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

16t^2+16t-60=0

a = 16; b = 16; c = -60;
Δ = b2-4ac
Δ = 162-4·16·(-60)
Δ = 4096
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4096}=64$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-64}{2*16}=\frac{-80}{32}
=-2+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+64}{2*16}=\frac{48}{32}
=1+1/2
$