6=t+t^2

Solution for 6=t+t^2 equation:

6=t+t^2

We move all terms to the left:

6-(t+t^2)=0

We get rid of parentheses

-t^2-t+6=0

We add all the numbers together, and all the variables

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

a = -1; b = -1; c = +6;
Δ = b2-4ac
Δ = -12-4·(-1)·6
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-5}{2*-1}=\frac{-4}{-2}
=+2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+5}{2*-1}=\frac{6}{-2}
=-3
$