t^2+2t-1680=0

Solution for t^2+2t-1680=0 equation:

t^2+2t-1680=0

a = 1; b = 2; c = -1680;
Δ = b2-4ac
Δ = 22-4·1·(-1680)
Δ = 6724
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6724}=82$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-82}{2*1}=\frac{-84}{2}
=-42
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+82}{2*1}=\frac{80}{2}
=40
$