t^2-2t-40=0

Solution for t^2-2t-40=0 equation:

t^2-2t-40=0

a = 1; b = -2; c = -40;
Δ = b2-4ac
Δ = -22-4·1·(-40)
Δ = 164
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:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{41}}{2*1}=\frac{2-2\sqrt{41}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{41}}{2*1}=\frac{2+2\sqrt{41}}{2}
$