5t^2-15t-5=0

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

5t^2-15t-5=0

a = 5; b = -15; c = -5;
Δ = b2-4ac
Δ = -152-4·5·(-5)
Δ = 325
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:

$\sqrt{\Delta}=\sqrt{325}=\sqrt{25*13}=\sqrt{25}*\sqrt{13}=5\sqrt{13}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-5\sqrt{13}}{2*5}=\frac{15-5\sqrt{13}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+5\sqrt{13}}{2*5}=\frac{15+5\sqrt{13}}{10}
$