t^2-12t-260=0

Solution for t^2-12t-260=0 equation:

t^2-12t-260=0

a = 1; b = -12; c = -260;
Δ = b2-4ac
Δ = -122-4·1·(-260)
Δ = 1184
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:

$\sqrt{\Delta}=\sqrt{1184}=\sqrt{16*74}=\sqrt{16}*\sqrt{74}=4\sqrt{74}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{74}}{2*1}=\frac{12-4\sqrt{74}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{74}}{2*1}=\frac{12+4\sqrt{74}}{2}
$