t^2+2t=500

Solution for t^2+2t=500 equation:

t^2+2t=500

We move all terms to the left:

t^2+2t-(500)=0

a = 1; b = 2; c = -500;
Δ = b2-4ac
Δ = 22-4·1·(-500)
Δ = 2004
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:

$\sqrt{\Delta}=\sqrt{2004}=\sqrt{4*501}=\sqrt{4}*\sqrt{501}=2\sqrt{501}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{501}}{2*1}=\frac{-2-2\sqrt{501}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{501}}{2*1}=\frac{-2+2\sqrt{501}}{2}
$