5t^2-25t-5=0

Solution for 5t^2-25t-5=0 equation:

5t^2-25t-5=0

a = 5; b = -25; c = -5;
Δ = b2-4ac
Δ = -252-4·5·(-5)
Δ = 725
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:

$\sqrt{\Delta}=\sqrt{725}=\sqrt{25*29}=\sqrt{25}*\sqrt{29}=5\sqrt{29}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-5\sqrt{29}}{2*5}=\frac{25-5\sqrt{29}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+5\sqrt{29}}{2*5}=\frac{25+5\sqrt{29}}{10}
$