16t^2+16t=480

Solution for 16t^2+16t=480 equation:

16t^2+16t=480

We move all terms to the left:

16t^2+16t-(480)=0

a = 16; b = 16; c = -480;
Δ = b2-4ac
Δ = 162-4·16·(-480)
Δ = 30976
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{30976}=176$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-176}{2*16}=\frac{-192}{32}
=-6
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+176}{2*16}=\frac{160}{32}
=5
$