16=t+2t^2

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

16=t+2t^2

We move all terms to the left:

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

We get rid of parentheses

-2t^2-t+16=0

We add all the numbers together, and all the variables

-2t^2-1t+16=0

a = -2; b = -1; c = +16;
Δ = b2-4ac
Δ = -12-4·(-2)·16
Δ = 129
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{129}}{2*-2}=\frac{1-\sqrt{129}}{-4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{129}}{2*-2}=\frac{1+\sqrt{129}}{-4}
$