10t^2-10t-2.5=0

Solution for 10t^2-10t-2.5=0 equation:

10t^2-10t-2.5=0

a = 10; b = -10; c = -2.5;
Δ = b2-4ac
Δ = -102-4·10·(-2.5)
Δ = 200
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:

$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10\sqrt{2}}{2*10}=\frac{10-10\sqrt{2}}{20}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10\sqrt{2}}{2*10}=\frac{10+10\sqrt{2}}{20}
$