16(t+3)(t-5)=0

Solution for 16(t+3)(t-5)=0 equation:

16(t+3)(t-5)=0

We multiply parentheses ..

16(+t^2-5t+3t-15)=0

We multiply parentheses

16t^2-80t+48t-240=0

We add all the numbers together, and all the variables

16t^2-32t-240=0

a = 16; b = -32; c = -240;
Δ = b2-4ac
Δ = -322-4·16·(-240)
Δ = 16384
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16384}=128$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-128}{2*16}=\frac{-96}{32}
=-3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+128}{2*16}=\frac{160}{32}
=5
$