-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}
$