5t^2-15t-1=0

Solution for 5t^2-15t-1=0 equation:

5t^2-15t-1=0

a = 5; b = -15; c = -1;
Δ = b2-4ac
Δ = -152-4·5·(-1)
Δ = 245
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:

$\sqrt{\Delta}=\sqrt{245}=\sqrt{49*5}=\sqrt{49}*\sqrt{5}=7\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-7\sqrt{5}}{2*5}=\frac{15-7\sqrt{5}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+7\sqrt{5}}{2*5}=\frac{15+7\sqrt{5}}{10}
$