6=2t+t^2

Solution for 6=2t+t^2 equation:

6=2t+t^2

We move all terms to the left:

6-(2t+t^2)=0

We get rid of parentheses

-t^2-2t+6=0

We add all the numbers together, and all the variables

-1t^2-2t+6=0

a = -1; b = -2; c = +6;
Δ = b2-4ac
Δ = -22-4·(-1)·6
Δ = 28
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{7}}{2*-1}=\frac{2-2\sqrt{7}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{7}}{2*-1}=\frac{2+2\sqrt{7}}{-2}
$