t^2+2t=6

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

t^2+2t=6

We move all terms to the left:

t^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}
$