0=t^2+6t+1

Solution for 0=t^2+6t+1 equation:

0=t^2+6t+1

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

-t^2-6t-1=0

We add all the numbers together, and all the variables

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

a = -1; b = -6; c = -1;
Δ = b2-4ac
Δ = -62-4·(-1)·(-1)
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-4\sqrt{2}}{2*-1}=\frac{6-4\sqrt{2}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+4\sqrt{2}}{2*-1}=\frac{6+4\sqrt{2}}{-2}
$