14=t^2+4t+2

Solution for 14=t^2+4t+2 equation:

14=t^2+4t+2

We move all terms to the left:

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

We get rid of parentheses

-t^2-4t-2+14=0

We add all the numbers together, and all the variables

-1t^2-4t+12=0

a = -1; b = -4; c = +12;
Δ = b2-4ac
Δ = -42-4·(-1)·12
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8}{2*-1}=\frac{-4}{-2}
=+2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*-1}=\frac{12}{-2}
=-6
$