t^2-2t-9=0

Solution for t^2-2t-9=0 equation:

t^2-2t-9=0

a = 1; b = -2; c = -9;
Δ = b2-4ac
Δ = -22-4·1·(-9)
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{10}}{2*1}=\frac{2-2\sqrt{10}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{10}}{2*1}=\frac{2+2\sqrt{10}}{2}
$