t^2-4t-260=0

Solution for t^2-4t-260=0 equation:

t^2-4t-260=0

a = 1; b = -4; c = -260;
Δ = b2-4ac
Δ = -42-4·1·(-260)
Δ = 1056
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:

$\sqrt{\Delta}=\sqrt{1056}=\sqrt{16*66}=\sqrt{16}*\sqrt{66}=4\sqrt{66}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{66}}{2*1}=\frac{4-4\sqrt{66}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{66}}{2*1}=\frac{4+4\sqrt{66}}{2}
$