16t^2+48t=160

Solution for 16t^2+48t=160 equation:

16t^2+48t=160

We move all terms to the left:

16t^2+48t-(160)=0

a = 16; b = 48; c = -160;
Δ = b2-4ac
Δ = 482-4·16·(-160)
Δ = 12544
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12544}=112$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-112}{2*16}=\frac{-160}{32}
=-5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+112}{2*16}=\frac{64}{32}
=2
$