6=t^2+8t

Solution for 6=t^2+8t equation:

6=t^2+8t

We move all terms to the left:

6-(t^2+8t)=0

We get rid of parentheses

-t^2-8t+6=0

We add all the numbers together, and all the variables

-1t^2-8t+6=0

a = -1; b = -8; c = +6;
Δ = b2-4ac
Δ = -82-4·(-1)·6
Δ = 88
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:

$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{22}}{2*-1}=\frac{8-2\sqrt{22}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{22}}{2*-1}=\frac{8+2\sqrt{22}}{-2}
$