16=(2t^2)+4t

Solution for 16=(2t^2)+4t equation:

16=(2t^2)+4t

We move all terms to the left:

16-((2t^2)+4t)=0

We get rid of parentheses

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

a = -2; b = -4; c = +16;
Δ = b2-4ac
Δ = -42-4·(-2)·16
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-12}{2*-2}=\frac{-8}{-4}
=+2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+12}{2*-2}=\frac{16}{-4}
=-4
$